This observation demonstrates the principle of electrostatics, which is the study of electric charges at rest. It highlights the fact that charged objects can interact with each other, and that opposite charges attract while like charges repel.
When a plastic rod is rubbed with cotton, electrons from the rod are transferred to the cotton, leaving the rod with an excess of positive charges, and the cotton with an excess of negative charges. This results in the rod acquiring a negative charge.
When the same cotton is brought near the cap of a positively charged electroscope, the leaves of the electroscope will diverge. This happens because the positively charged cap attracts the negative charges on the cotton, causing the electrons to move towards the cap. As a result, the leaves of the electroscope acquire a negative charge, and they repel each other due to their like charges.
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A cosmetic mirror is designed to magnify your face by a factor of 1.32 when your face is 19.5 cm in front of it.
a. Calculate the radius of curvature of the mirror (in cm)
The radius of curvature of the mirror is 761.9 cm or 7.619 meters.
How to determine radius?Assuming the mirror is a spherical mirror, use the mirror equation:
1/f = 1/di + 1/do
where f = focal length, di = image distance (distance from the mirror to the image), and do = object distance (distance from the mirror to the object).
Since the mirror magnifies the face by a factor of 1.32, the image distance is 1.32 times the object distance:
di = 1.32 × do
The object distance is given as 19.5 cm, so substitute to get:
di = 1.32 × 19.5 cm = 25.74 cm
The mirror magnifies the face by a factor of 1.32, so the magnification is:
m = -di/do = 1.32
Since the mirror is concave (it magnifies the image), the magnification is negative. Substituting di and do:
-1.32 = -25.74 cm/do
Solving for do:
do = 19.5 cm × 25.74 cm / 1.32 / 25.74 cm = 380.95 cm
The object distance is the distance from the mirror to the object, which is half the radius of curvature (R) of the mirror:
do = R/2
So we can solve for R:
R = 2 × do = 2 × 380.95 cm = 761.9 cm
Therefore, the radius of curvature of the mirror is 761.9 cm or 7.619 meters.
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a walker can see a cliff face some distance away and wants to estimate how far away it is. He makes a loud noise and times how long it takes the echo to come back. It takes 1.8s if the speed of sound is 345 m/s how far away is the cliff?
The cliff is approximately 621 meters away from the walker.
To determine the distance to the cliff, we can use the fact that sound travels at a constant speed through the air. By measuring the time it takes for the echo to return, we can calculate the distance traveled by the sound wave.Given that the time taken for the echo to come back is 1.8 seconds and the speed of sound is 345 m/s, we can use the formula: distance = speed × time.
Therefore, the distance to the cliff can be calculated as follows:
Distance = Speed of sound × Time
= 345 m/s × 1.8 s
= 621 meters.
This calculation assumes that the sound waves traveled in a straight line to the cliff and back, without any significant obstructions or reflections. It's important to note that this method provides an estimation of the distance, as environmental factors like wind and temperature can affect the speed of sound.
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the area of a larger piston in a hydraulic press is 4 m squared and that of the other person is 0.05 m squared a force of 100 Newton is applied on the smaller piston how much force is produced than the larger piston
The force produced on the larger piston is 8000 N.
What is force?
Force is the product of mass and acceleration
To calculate the force produced in the larger piston, we use the formula below.
Formula:
F/A = f/a......................... Equation 1Where:
F = Force on the larger pistonA = Area of the larger pistonf = Force on the smaller pistona = Area of the smaller pistonFrom the question,
Given:
f = 100 Na = 0.05 m²A = 4 m²Substitute thee values into equation 1 and solve for F
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A proton moves across the Earth's equator in a northeasterly direction. At this point the earth's magnetic field has a direction due north and is parallel to the surface. What is the direction of the force acting on the proton at this instant?
Answer:
the direction is Southly west
/ Assignment 01 Science teachers think practical science is a good thing. a. Agree b. Disagree c. Uncertain
It is reasonable to conclude that science teachers generally agree that practical science is beneficial. Therefore, option (A) is correct.
Practical science refers to hands-on activities, experiments, and applications of scientific concepts in real-world settings. Science teachers, who are experts in their field and experienced in teaching science, understand the importance of practical science in facilitating students' understanding, engagement, and application of scientific principles.
Practical science allows students to develop critical thinking, problem-solving, and inquiry skills, as well as promoting a deeper understanding of scientific concepts. It also fosters curiosity, creativity, and a passion for science, making it an effective and essential component of science education.
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The coefficient of static friction between a 20 kg weight and a football turf is .85. What force is needed to make the weight start moving
A force of 166.8 N is needed to make the weight start moving.
The force needed to make the 20 kg weight start moving can be calculated using the formula F = μsN,
where F is the force required to overcome static friction,
μs is the coefficient of static friction,
and N is the normal force exerted on the weight by the turf.
The normal force is equal to the weight of the object,
which is 20 kg multiplied by the acceleration due to gravity, 9.81 [tex]m/s^2[/tex],
so N = 196.2 N.
Substituting the given values into the formula,
we have F = 0.85 × 196.2 N = 166.8 N.
Static friction is the force that prevents an object from moving when it is at rest on a surface. The coefficient of static friction is a measure of the grip between the two surfaces, with higher values indicating greater friction. In this case, the high coefficient of static friction between the weight and the football turf means that a relatively large force is needed to overcome the friction and start the weight moving.
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Discuss the consequences of poor health and safety when using non-ionising and ionising radiation technologies and the prevention and safety measures employed.
Poor health and safety practices when using non-ionizing and ionizing radiation technologies can have significant consequences. Here are some potential consequences and the prevention and safety measures employed to mitigate them.
Consequences of poor health and safety:
Non-Ionizing Radiation:
Thermal Effects: Prolonged exposure to high levels of non-ionizing radiation, such as from radiofrequency waves (RF) or microwaves, can lead to thermal effects, including tissue heating and burns.Ionizing Radiation:
Acute Radiation Syndrome (ARS): High doses of ionizing radiation, such as those experienced during a nuclear accident or radiation therapy accidents, can lead to ARS, which includes symptoms like nausea, vomiting, fatigue, and potentially fatal damage to vital organs.Prevention and safety measures:
Non-Ionizing Radiation:
Time, Distance, and Shielding: Minimize exposure time to non-ionizing radiation sources, maintain a safe distance from the source, and use shielding materials (e.g., lead aprons, and protective barriers) to reduce exposure.Compliance with Guidelines: Follow national and international guidelines and regulations regarding safe exposure limits for different types of non-ionizing radiation sources. For example, adherence to specific absorption rate (SAR) limits for mobile devices.Ionizing Radiation:
Risk Assessment and Minimization: Conduct thorough risk assessments to identify potential hazards, establish safety protocols, and minimize radiation exposure through engineering controls and administrative measures.Personal Protective Equipment (PPE): Use appropriate PPE, such as lead aprons, gloves, and goggles, to protect against direct radiation exposure.Training and Education: Provide comprehensive training and education to workers who handle ionizing radiation sources to ensure they are aware of the risks, safety protocols, and proper handling procedures.Monitoring and Dosimetry: Regularly monitor radiation levels in the workplace and ensure that employees wear dosimeters to measure their radiation exposure.Regulatory Compliance: Adhere to applicable regulations and standards, such as those set by radiation protection agencies and authorities, to ensure compliance with safety measures and limits on radiation exposure.know more about ionizing radiation here:
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Question 2 2.1 Two resistors with values of 22 and 6 are connected in parallel. This combination is then connected in series with a 3 resistor. The supply voltage of the whole circuit is 12 V. 2.1.1 Draw a neat, labelled diagram of the circuit. 2.1.2 Calculate the following: a) the resistance of the parallel pair of resistors b) the total resistance of the circuit c) the total current flow of the circuit d) the voltage drop across each resistor e) the current flowing through the 2 f) the current flowing through the 6 116 Topic 3 Direct current (DC) and alternating current (AC) circuits resistor resistor.
a. the resistance of the parallel pair of resistors is 4.71Ω.
b. the total resistance of the circuit is 7.71Ω.
c. the total current flow of the circuit is 1.55A.
d. The voltage drop across each resistor is 12V.
e. he current flowing through the 2Ω resistor is 1.55A.
f. the current flowing through the 6Ω resistor is 1.55A.
How do we calculate?Equivalent resistance :
1/Req = 1/R1 + 1/R2
1/Req = 1/22Ω + 1/6Ω
1/Req = (6 + 22)/(22 * 6)
1/Req = 28/132
Equivalent resistance = 132/28
Equivalent resistance = 4.71Ω
b) The total resistance of the circuit:
total resistance = equivalent resistance + R3
total resistance = 4.71Ω + 3Ω
total resistance = 7.71Ω
c) The total current flow of the circuit:
We use Ohm's law
I = V / R
I = 12V / 7.71Ω
I = 1.55A
d) The voltage drop across each resistor is the same as the total voltage
e) The current flowing through the 2Ω resistor is same as all resistors.
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derivation of green's function for schrodinger wave equation?
The derivation of Green's function for the Schrödinger wave equation involves using the concept of superposition of solutions to find a particular solution for a given initial condition.
Green's function is a fine tool that's used to break discriminational equations, including the Schrödinger equation. The idea behind Green's function is to find a result to the discriminational equation that satisfies a given original condition. In the environment of the Schrödinger equation, Green's function represents the probability breadth of chancing a flyspeck at a particular position at a particular time, given that it started from a known original position and time.
The Green's function is attained by working the Schrödinger equation for a delta function source at the original position. To decide Green's function, one can consider the Schrödinger equation with a delta function source at some original position. This leads to the result for the surge function as a sum of two terms- a free flyspeck term and a term due to the delta function source.
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1. A student drives 40 km south in an attempt to get to a science fair competition before realizing he missed the exit. He turns
around and drives back north 15 km before arriving. The total time of travel was 0.75 hr. What is his speed? What is his
velocity?
2. A bus leaves Houston at 6:00 am headed toward Austin at a constant speed of 54.3 miles/hour. If Austin is located 190 miles west of Houston, approximately when will the bus arrive in Austin?
3. A girl rides her moped 5 km east and then turns around and rides 2 km west. The entire trip takes fifteen minutes. What are
her average speed and velocity?
Answer:
1) = approximately 74 km/hr
2) = 9:30 AM
3) Avg speed = 28 km/hr
vel = 0.
Explanation:
Here is how you do it:
1) To find the student's speed, we can use the formula:
Speed = Distance / Time
The student traveled a total distance of 40 km south and 15 km north, which gives a total distance of 40 km + 15 km = 55 km. The total time of travel was 0.75 hours.
Speed = 55 km / 0.75 hr ≈ 73.33 km/hr
Therefore, the student's speed is approximately 73.33 km/hr or 74 km/hr.
To calculate the student's velocity, we need to consider both magnitude and direction. Since the student ended up at the same position as the starting point, the displacement is zero. Therefore, the velocity is also zero.
2) The distance between Houston and Austin is 190 miles. The bus is traveling at a constant speed of 54.3 miles/hour. We can use the formula:
Time = Distance / Speed
Time = 190 miles / 54.3 miles/hour ≈ 3.50 hours
The bus will arrive in Austin approximately 3.50 hours after leaving Houston.
If the bus leaves Houston at 6:00 am, it will arrive in Austin around 9:30 am.
3) The girl rode 5 km east and then turned around and rode 2 km west. The total distance traveled is 5 km + 2 km = 7 km. The entire trip took fifteen minutes, which is equal to 15/60 = 0.25 hours.
Average Speed = Total Distance / Total Time
Average Speed = 7 km / 0.25 hr = 28 km/hr
Therefore, the girl's average speed is 28 km/hr.
To calculate the girl's velocity, we need to consider both magnitude and direction. The girl's initial direction was east, and her final direction was west. Since the starting and ending points are the same, the displacement is zero. Therefore, the velocity is also zero.
a ball is thrown directly downward with an initial speed of 8.45 m/s, from a height of 29.8m. After what time interval does it strike the ground?
Answer:
Approximately [tex]1.75\; {\rm s}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance is negligible.)
Explanation:
Find the velocity of the ball right before landing using the following SUVAT equation:
[tex]\displaystyle v^{2} - u^{2} = 2\, a\, x[/tex],
Where:
[tex]v[/tex] is the velocity of the ball right before landing,[tex]u = 8.45\; {\rm m\cdot s^{-1}}[/tex] is the initial velocity,[tex]a = 9.81\; {\rm m\cdot s^{-2}}[/tex] is the acceleration, and[tex]x = 29.8\; {\rm m}[/tex] is the change in the height of the ball.Rearrange this equation to find [tex]v[/tex]:
[tex]\begin{aligned}v &= \sqrt{u^{2} + 2\, a\, x} \\ &= \sqrt{(8.45)^{2} + 2\, (9.81)\, (29.8)}\; {\rm m\cdot s^{-1}} \\ &\approx 25.61\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Divide the change in velocity by acceleration to find the time elapsed:
[tex]\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{25.61 - 8.45}{9.81} \; {\rm s}\\ &\approx 1.75\; {\rm s}\end{aligned}[/tex].
a sound wave traveling through a certain freshwater lake has a frequency of 257.2hz and a wavelength of 3.25m. if the water conditions are held constant, all sound waves will travel at the save speed through water. use this fact to calculate the wavelength of a sound wave with a frequency of 415.3hz. (show work pls<3)
The wavelength of the sound wave with a frequency of 415.3 Hz is approximately 3.565 m.
How to determine wavelength?The speed of sound in water is approximately 1482 m/s. Use the formula v = f × λ, where v = velocity (speed of sound), f = frequency, and λ = wavelength.
Given:
Frequency of the first sound wave (f₁) = 257.2 Hz
Wavelength of the first sound wave (λ₁) = 3.25 m
Velocity of sound in water (v) = 1482 m/s
Rearrange the formula to solve for λ₂ (wavelength of the second sound wave):
v = f × λ
λ = v / f
Substituting the values:
λ₂ = v / f₂
= 1482 m/s / 415.3 Hz
Calculating:
λ₂ ≈ 3.565 m
Therefore, the wavelength of the sound wave with a frequency of 415.3 Hz is approximately 3.565 m.
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A machine part consists of three heavy disks linked by struts of negligible weights as shown in the figure. Calculate the moment of inertia of the body about an axis through the centre of disk A and the kinetic energy, if the body rotates about an axis through A perpendicular to the plane of the diagram, with angular speed ω = 6.0 rads-1..
if the values of mass (m) and radius (R) are provided, the moment of inertia of the body about an axis through the center of disk A can be calculated as (3/2) * m * R^2, and the kinetic energy of the rotating body would be 162 * m * R^2 Joules.
To calculate the moment of inertia of the body about an axis through the center of disk A, we need to consider the moment of inertia contributions from each individual disk and add them up.
Let's denote the moment of inertia of each disk as I_A, I_B, and I_C, respectively. The moment of inertia of a disk rotating about its center can be calculated using the formula:
I = (1/2) * m * r^2
Where m is the mass of the disk and r is its radius.
Since the struts have negligible weight, we can assume that each disk has the same mass.
Let's assume the mass of each disk is m and the radius of each disk is R.
The moment of inertia of disk A (I_A) is given by:
I_A = (1/2) * m * R^2
The moment of inertia of disk B (I_B) and disk C (I_C) will be the same since they have the same mass and radius:
I_B = I_C = (1/2) * m * R^2
The total moment of inertia of the body about the axis through the center of disk A (I_total) is the sum of the individual moment of inertias:
I_total = I_A + I_B + I_C
= (1/2) * m * R^2 + (1/2) * m * R^2 + (1/2) * m * R^2
= (3/2) * m * R^2
To calculate the kinetic energy of the rotating body, we can use the formula:
Kinetic Energy = (1/2) * I_total * ω^2
Substituting the given values:
Kinetic Energy = (1/2) * ((3/2) * m * R^2) * (6.0 rad/s)^2
Simplifying further, if the values of m and R are given, we can calculate the moment of inertia and kinetic energy.
Assuming that the values of mass (m) and radius (R) are given, we can calculate the moment of inertia (I_total) and kinetic energy.
For the given values of ω = 6.0 rad/s and the previously calculated I_total:
I_total = (3/2) * m * R^2
Kinetic Energy = (1/2) * I_total * ω^2
= (1/2) * [(3/2) * m * R^2] * (6.0 rad/s)^2
= (9/2) * m * R^2 * (36.0 rad^2/s^2)
= 162 * m * R^2 Joules
Therefore, if the values of mass (m) and radius (R) are provided, the moment of inertia of the body about an axis through the center of disk A can be calculated as (3/2) * m * R^2, and the kinetic energy of the rotating body would be 162 * m * R^2 Joules.
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Now, the rock is at the bottom of the cliff, just before touching the
ground.
250 m
10 kg
a. What is the rock's gravitational potential energy, just before
touching the ground? 0
b. What is the rock's kinetic energy, just before touching the
ground?
a. The rock's gravitational potential energy, just before touching the ground is 0 J.
b. The rock's kinetic energy, just before touching the ground is 24,500 J.
What is rock's gravitational potential energy?The rock's gravitational potential energy, just before touching the ground is calculated as follows;
Just before touching the ground, the rock will have maximum velocity, and the kinetic energy of rock will be maximum while the gravitational potential energy will be minimum or zero.
P.E (top height) = K.E (minimum or zero height)
P.E = mgh
where;
m is the massg is gravityh is heightP.E = 10 x 9.8 x 250
P.E = 24,500 J
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Question 2 The pulley on a machine is 230 mm diameter. It is to be driven at 183 rev/min. A main shaft to drive the machine has a pulley of diameter 140 mm. What is the speed of the running shaft driving the machine? (10)
The speed of the running shaft driving the machine, given that the main shaft to drive the machine has a pulley of diameter 140 mm is 111.4 rev/min
How do i determine the speed of the running shaft?First, we shall list out the given parameters from the question. Details below:
Speed of the main shaft (S₁) = 183 rev/minDiameter of the main shaft (D₁) = 140 mmDiameter of the second pulley (D₂) = 230 mmSpeed of the running shaft i.e second pulley (S₂) = ?The speed of the running shaft can be obtain as shown below:
S₁D₁ = S₂D₂
183 × 140 = S₂ × 230
183 × 140 = S₂ × 230
25620 = S₂ × 230
Divide both sides by 230
S₂ = 25620 / 230
S₂ = 111.4 rev/min
Thus, we can conclude that the speed of the running shaft driving the machine is 111.4 rev/min
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two mass damper on a lagos tower consist of a 373mg concrete block.that complete one oscillation in.6.80 secs .the oscillation amplitude in a high wind is 110cm. determine the spring constant
If two mass dampers on a Lagos tower consist of a 373mg concrete block.that complete one oscillation in.6.80 secs. the oscillation amplitude in a high wind is 110cm. Then the spring constant of the mass-spring system is 6.90 N/m.
The spring constant is a measure of the stiffness of a spring, which describes how much force is required to stretch or compress the spring by a certain amount. It is typically measured in units of newtons per meter (N/m).
We can use the formula for the period of oscillation of a mass-spring system:
T = 2π√(m/k)
where T is the period of oscillation, m is the mass of the block, and k is the spring constant.
We can rearrange this formula to solve for k:
k = (4π²m) / T²
where we substitute the values given in the problem.
m = 373mg = 0.373g = 0.000373kg (convert milligrams to kilograms)
T = 6.80 s
π ≈ 3.14159
Plugging in these values, we get:
k = (4π² × 0.000373) / (6.80)²
≈ 6.90 N/m
Therefore, the spring constant of the mass-spring system is approximately 6.90 N/m.
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Marble M moves to the right before it strikes marble N, which is initially at rest. Both marbles have the same mass, 4.1 g. After the collision, the marbles move in the directions shown in the diagram. Marble M moves away from the collision point with a speed of 23.9 cm/s.
What is the initial kinetic energy of marble M?
6.1 × 103 g . cm/s2
8.3 × 103 g . cm/s2
9.4 × 103 g . cm/s2
7.2 × 103 g . cm/s2
The initial kinetic energy of marble M based on the information is D. 7.2 × [tex]10^{3}[/tex]gcm/s²
How to explain the informationSubstituting the masses and velocities from the problem statement, we get:
K₂ = 1/2 * 8.2 g * (23.9 cm/s)²
K₂ = 8.3 × [tex]10^{3}[/tex] g/cm/s²
p₂ = 4.1 g * ((m₁/m₂) * v3 + v₃)
p₂ = 4.1 g * (1 + m₁/m₂) * v₃
Since p1 = p2, we can equate the expressions for p₁ and p₂ and solve for v₃:
m₁ * v1 = 4.1 g * (1 + m₁/m₂) * v₃
v₃ = (m1 * v1) / (4.1 g * (1 + m₁/m₂))
Substituting the masses and velocities from the problem statement, we get:
v3 = (4.1 g * 23.9 cm/s) / (4.1 g * 2)
v3 = 11.95 cm/s
K₁ = 1/2 * 4.1 g * (23.9 cm/s)²
K₁ = 7.2 × [tex]10^{3}[/tex]g/cm/s²
Therefore, the initial kinetic energy of marble M is 7.2 × 10³g cm/s². Thus, the correct option is 7.2 × [tex]10^{3}[/tex] g cm/s²
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what type of path do people in the plane observe that the pack follows?
Answer:
As can be seen from the above animation, the package follows a parabolic path and remains directly below the plane at all times.
Explanation:
hope it helps
Q3. The curvilinear motion of a particle is defined by v, =50-16t and y=100-4t', where v, is in meters per second, y is in meters, and t is in seconds. It is also known that x=0 when t=0.Plot the path of the particle and determine its velocity and acceleration when the position y=0 is reached, Write the vector representation of the answer.
The velocity is [tex]10\hat{i}-16\hat{j} m/s[/tex] and the acceleration is [tex]-16\hat{i} -8\hat{j} m/s^2[/tex]
The given equations are:
[tex]v_x = 50 - 16t\\y = 100 - 4t^2[/tex]
x = 0, when t = 0
To plot the path of the particle, we need to eliminate t from the given equations and express y in terms of x. Using the given information, we have:
[tex]t = (50 - v_x) / 16\\y = 100 - 4t^2 = 100 - 4[(50 - v_x) / 16]^2[/tex]
Simplifying the above equation, we get:
[tex]y = (25/2) - (5/4)v_x + (1/32)v_x^2[/tex]
This is the equation of the path followed by the particle.
To determine the velocity and acceleration when y = 0, we need to find the corresponding value of [tex]v_x[/tex]. Setting y = 0 in the above equation, we get:
[tex]v_x = 20 or v_x = 10[/tex]
When v_x = 20, the particle is moving in the positive x-direction. The velocity and acceleration can be calculated as follows:
Velocity =[tex](v_x, v_y) = (20, -32)[/tex]
Acceleration = [tex](a_x, a_y) = (-16, -8)[/tex]
When [tex]v_x = 10[/tex], the particle is moving in the negative x-direction. The velocity and acceleration can be calculated as follows:
Velocity =[tex](v_x, v_y) = (10, -16)[/tex]
Acceleration = [tex](a_x, a_y) = (-16, -8)[/tex]
Therefore, the vector representation of the velocity and acceleration when y = 0 are:
For[tex]v_x = 20[/tex]: Velocity = (20, -32) m/s; Acceleration = (-16, -8) [tex]m/s^2[/tex]
For[tex]v_x = 10[/tex]: Velocity = (10, -16) m/s; Acceleration = (-16, -8)[tex]m/s^2[/tex]
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Particles q1= -66.3 μC, q2 = +108 μC, and q3 = -43.2 μC are in a line. Particles q1 and q2 are separated by 0.550 m and particles q2 and q3 are separated by 0.550 m. What is the net force on particle q2?
Remember:
Negative forces (-F) will point Left
Positive forces (+F) will point Right
Will mark brainliest IF answer is correct.
To calculate the net force on particle q2, we need to calculate the individual forces exerted by q1 and q3 on q2, considering their charges and separation distances.
Using Coulomb's Law, the force between two charged particles can be calculated as:
[tex]F = (k * |q1 * q2|) / r^2[/tex]
where F is the force, k is Coulomb's constant (8.99 x 10^9 N m²/C²), q1 and q2 are the charges of the particles, and r is the separation distance between them.
For q1 and q2:
F1 = (8.99 x 10^9 N m²/C²) * (|-66.3 x 10^-6 C * 108 x 10^-6 C|) / (0.550 m)²
For q2 and q3:
F2 = (8.99 x 10^9 N m²/C²) * (|108 x 10^-6 C * -43.2 x 10^-6 C|) / (0.550 m)²
The net force on q2 is the vector sum of F1 and F2:
Net Force = F1 + F2
By calculating these values and performing the addition, we can determine the net force acting on particle q2.
Therefore, To calculate the net force on particle q2, we need to calculate the individual forces exerted by q1 and q3 on q2, considering their charges and separation distances.
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A bus with a maximum speed of 20m/s takes 21sec to travel 270m from stop to stop. Its acceleration is twice as great as its deceleration.
Find
1. The acceleration
2. The distance travelled at maximum speed
The acceleration can be calculated using the formula:
Acceleration = (Final velocity - Initial velocity) / Time taken
Given that the bus starts from rest, the initial velocity is 0 m/s.
Acceleration = (20 m/s - 0 m/s) / 21 sec = 20/21 m/s².
The distance travelled at maximum speed can be calculated by subtracting the distances covered during acceleration and deceleration from the total distance.
Distance during acceleration = (1/2) * acceleration * time² = (1/2) * (20/21 m/s²) * (21 sec)² = 210 m.
Distance during deceleration is the same as distance during acceleration.
Distance travelled at maximum speed = Total distance - 2 * distance during acceleration = 270 m - 2 * 210 m = -150 m.
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In a 2X4X3 factorial design, how many levels does the second independent variable have?
Answer:
The second independent variable in a 2x4x3 factorial design has 4 levels.
Explanation:
Hi please answer the question labeled d
If Q1 is 5 times larger than Q2 the force that Q1 exerts on Q2 is?
(greater than, smaller than, or exactly the same as) the force that Q2 exerts on Q1.
part a) If Q increases by 5 times its original value, the electrostatic force (F) will increase5 times as well.
part b) If r is halved (reduced by 2), the force will become four times stronger (since 2² = 4).
part c) If Q1 is positive and Q2 is negative, the charges will attract each other.
part d) If the force that Q1 exerts on Q2 is 5 times larger than the force that Q2 exerts on Q1 is same.
What is electrostatic force?The electrostatic force is described as the force of attraction or repulsion between two charged particles.
With regards to Coulomb's Law, we have that the electrostatic force between two charges separated by a distance is :
Force = k * (Q1 * Q2) / r²
Where:
F_ = electrostatic force
k = electrostatic constant
Q1 and Q2 = magnitudes of the charges
r = distance
for case a:
If one of the charges, Q1 or Q2, increases by 5 times then the electrostatic force will also increase by 5 times.
case b)
If the distance between the charges, r, is halved, the electrostatic force will become four times stronger because (1/r²).
for case c.
if Q1 is positive and Q2 is negative, the charges will attract each other because of magnetic laws.
for case d.)
If the force that Q1 exerts on Q2 is 5 times larger than the force that Q2 exerts on Q1 is same as there is a resulting stronger gravitational or electromagnetic force.
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Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The
object moves with constant acceleration. Which part of a velocity vs. time graph can be used to calculate the
displacement of the object?
O the area of the rectangle under the line
the area of the rectangle above the line
the area of the rectangle plus the area of the triangle under the line
the area of the rectangle plus the area of the triangle above the line
Save and Exit
Next
Submit
Answer: The correct answer is:
the area of the trapezoid under the line
To explain this, let's consider the velocity vs. time graph again. Since the object moves with constant acceleration, the graph will be a straight line with a positive slope. The area under the line represents the distance traveled by the object, which is equal to the displacement if the initial position is zero.
The area under the line is a trapezoid because the velocity is changing over time. The base of the trapezoid is the time interval, and the heights are the initial and final velocities. The formula for the area of a trapezoid is:
Area = (base1 + base2) / 2 * height
where:
base1 = initial velocity
base2 = final velocity
height = time interval
Substituting the given values, we get:
Area = (v_i + v_f) / 2 * t
where v_i = 3 m/s, v_f = 10 m/s, and t is the time interval over which the velocities change.
Therefore, the correct answer is the area of the trapezoid under the line.
The force between two charged objects is 200N. If. the charge of one object increaes 3x and the other charge decreaes 4x, what will the new force be?
Given that the force between two charged particles, we'll call charge 1 and 2 is, [tex]\vec F_{0}=200 \ N[/tex]. The question asks us to find the new force between charges 1 and 2 if the charge on 1 increases by 3 times and the charge on 2 decreases by 4 times.
Equation to calculate the electric force between two charged particles:
[tex]\vec F_e=k_e\frac{q_1q_2}{r^2} \\\\ k_e=Coulomb's \ Constant= 8.99 \times 10^9 \ \frac{Nm^2}{C^2}\\[/tex]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
[tex]\vec F_0=k_e\frac{q_1q_2}{r^2} =200 \ N \ and \ \boxed{\vec F_f=k_e\frac{(3 q_1)(\frac{1}{4} q_2)}{r^2} }[/tex]
[tex]\Longrightarrow \vec F_f=k_e\frac{(3 q_1)(\frac{1}{4} q_2)}{r^2} \Longrightarrow \vec F_f=\frac{3}{4} k_e\frac{q_1 q_2}{r^2} \Longrightarrow \vec F_f=\frac{3}{4}(200) \Longrightarrow \boxed{\boxed{\vec F_f=150 \ N}}[/tex]
Thus, the new force would be 150 N.
The new force will be 150N
The force between two charged objects is given by Coulomb's law, which states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, Coulomb's law can be expressed as:
F = k * (q1 * q2) / r^2
where F is the force, k is Coulomb's constant, q1 and q2 are the charges of the two objects, and r is the distance between them.
In this problem, we are given that the force between the two objects is 200N. Let us assume that the charges of the two objects are q1 and q2, respectively. Using Coulomb's law, we can write:
200 = k * (q1 * q2) / r^2
Now, we are told that the charge of one object increases by a factor of 3, and the charge of the other object decreases by a factor of 4. Let us call the new charges q1' and q2', respectively. We can write:
q1' = 3 * q1
q2' = (1/4) * q2
Substituting these expressions into Coulomb's law, we get:
F' = k * (q1' * q2') / r^2
= k * [(3 * q1) * ((1/4) * q2)] / r^2
= k * (3/4) * (q1 * q2) / r^2
= (3/4) * F
Therefore, the new force between the two charged objects is 3/4 times the original force, or 150N.
In summary, the force between two charged objects is proportional to the product of the charges and inversely proportional to the square of the distance between them. If one charge increases by a factor of 3 and the other charge decreases by a factor of 4, the new force between them will be 3/4 times the original force. This is because the product of the charges is multiplied by (3/4) in Coulomb's law.
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suppose you are given common salt and edible oil if you have to carry out the comparative of their properties with scientific learning will you use
Answer:
Sure, I can help you with that. Here are some of the properties of common salt and edible oil that can be compared using scientific learning:
Physical properties: Common salt is a solid at room temperature, while edible oil is a liquid. Common salt is also denser than edible oil.
Chemical properties: Common salt is a compound of sodium and chlorine, while edible oil is a mixture of triglycerides.
Biological properties: Common salt is essential for human health, while edible oil is a source of energy and essential fatty acids.
To compare the properties of common salt and edible oil, I would use the following scientific methods:
Observation: I would observe the physical properties of common salt and edible oil, such as their color, odor, and texture.
Measurement: I would measure the physical properties of common salt and edible oil, such as their melting point, boiling point, and density.
Chemical analysis: I would use chemical analysis to determine the chemical composition of common salt and edible oil.
Biological testing: I would use biological testing to determine the biological effects of common salt and edible oil on humans and animals.
By using these scientific methods, I would be able to compare the properties of common salt and edible oil in a comprehensive and informative way.
Explanation:
Common salt and edible oil can be compared using scientific learning by examining their physical and chemical properties. Through experiments and analysis, we can better understand the similarities and differences between them.
Explanation:Common salt (sodium chloride) and edible oil are both food ingredients, but they have distinct properties and serve different purposes in cooking and nutrition. The properties of common salt and edible oil can be compared using scientific learning in various ways. One approach is to examine their physical properties such as melting point, boiling point, solubility, and density.
Another approach is to analyze their chemical properties, including their reactions with other substances and their ability to conduct electricity. By conducting experiments and analyzing the data, we can gain a better understanding of the similarities and differences between common salt and edible oil. Edible oils provide essential fats and calories, while salt is used sparingly due to its association with health concerns like hypertension.
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A beam of light passes from air (n=1.0) into water (n = 1.33). If the angle of incidence is 77º, what is the
angle of refraction in the water?
A 33.36°
B 47.11°
C 0.97⁰
D 0.55⁰
Answer:
The correct answer is option B: 47.11°.
Explanation:
To calculate the angle of refraction when light passes from air to water, we can use Snell's law, which states:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
where:
n₁ = refractive index of the medium of incidence (in this case, air) = 1.0
n₂ = refractive index of the medium of refraction (in this case, water) = 1.33
θ₁ = angle of incidence
θ₂ = angle of refraction (what we want to find)
Given:
θ₁ = 77º
Let's plug in the values into Snell's law and solve for θ₂:
1.0 * sin(77º) = 1.33 * sin(θ₂)
sin(77º) = (1.33 * sin(θ₂)) / 1.0
Now, isolate sin(θ₂) by multiplying both sides by 1.0:
sin(77º) * 1.0 = 1.33 * sin(θ₂)
sin(θ₂) = (sin(77º) * 1.0) / 1.33
Now, take the inverse sine (arcsin) of both sides to find θ₂:
θ₂ = arcsin((sin(77º) * 1.0) / 1.33)
Calculating the value:
θ₂ ≈ 47.11º
Therefore, the angle of refraction in the water when the angle of incidence is 77º is approximately 47.11º.
Exercise II: A ball (C) of small dimensions of mass m=2kg is launched from A without initial speed, the force of friction exerted on (C) between A and B is f-IN. Take g=10m/s. (C) 0-30° B a) Applying Newton's second law between A and B, determine the nature of motion and the value of AB such that V 3m/s. b) Between B and C the force of friction fexerted on (C) is supposed to be constant. Determine its value such that Vc=2m/s and BC-2.5m. GOODWORK C
A) The nature of motion is decelerating motion and the value of AB = 5m. B) Between B and C the force of friction fexerted on 2.5m and the force of friction exerted on (C) between B and C is 17.6N
(a) Applying Newton's second law between A and B, determine the nature of motion and the value of AB such that V=3m/s.
From A to B, the ball experiences a force of friction and the gravitational force acting downwards. The force of friction opposes the motion of the ball, while the gravitational force acts in the direction of motion.
Since the ball is launched from A without any initial speed, it will initially move downwards due to the force of gravity until it reaches its lowest point at B. To determine the value of AB such that V=3m/s, we can use the conservation of energy principle.
The potential energy at A is zero, and the kinetic energy at B is (1/2)mv^2, where m is the mass of the ball and v is the velocity at B. The work done by the force of friction is fABd, where d is the distance from A to B. The work done by the gravitational force is -mg(h/2), where h is the height difference between A and B. Thus, we have (1/2)mv^2 = fABd - mg(h/2).
Since the ball is moving upwards at B, the net force acting on it is in the upward direction. Using Newton's second law, we have fAB - mg = ma, where a is the acceleration of the ball. Since the ball is moving upwards, a = -g. Solving these equations simultaneously, we obtain AB = 5m and the nature of motion is a decelerating motion.
(b) Between B and C, the force of friction f exerted on (C) is supposed to be constant. Determine its value such that Vc=2m/s and BC=2.5m.
Between B and C, the ball moves upwards due to the force of friction and the gravitational force acting downwards. To determine the force of friction f, we can use the conservation of energy principle again. The potential energy at B is mgh/2, where h is the height difference between A and B.
The kinetic energy at C is (1/2)mv^2, where v is the velocity at C. The work done by the force of friction is fBCd, where d is the distance from B to C. Thus, we have (1/2)mv^2 = fBCd + mgh/2. Using Newton's second law, we have fBC - mg = ma, where a is the acceleration of the ball. Since the ball is moving upwards, a = -g. Solving these equations simultaneously and substituting the given values, we obtain f = 17.6N. Therefore, the force of friction exerted on (C) between B and C is 17.6N.
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The coefficient of static friction between a 20 kg weight and a football turf is .85. What force is needed to make the weight start moving
The coefficient of static friction between a 20 kg weight and a football turf is .85. Force of approximately 166.6 Newtons (N) is needed to make the weight start moving.
The force needed to make the 20 kg weight start moving can be determined using the coefficient of static friction. The equation for static friction is:
Frictional force = coefficient of static friction * normal force
The normal force is the force exerted by the surface perpendicular to the weight. In this case, it is equal to the weight of the object, which can be calculated as the mass (20 kg) multiplied by the acceleration due to gravity (9.8 m/s^2):
Normal force = mass * gravity = 20 kg * 9.8 m/s^2 = 196 N
Now, we can calculate the force needed to make the weight start moving:
Frictional force = 0.85 * 196 N ≈ 166.6 N
Therefore, a force of approximately 166.6 Newtons (N) is needed to overcome the static friction and make the 20 kg weight start moving on the football turf. This force must be applied in the opposite direction to the frictional force to initiate motion.
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Use the prompt below to answer questions #20-22.
A 5.6kg uniform cylindrical grinding wheel has a radius of 0.28m. It is initially rotating at 30 rad/s. The
rotational inertia of the wheel is the same as for a solid disk, MR¹/2.
1-MRO
Axis
Solid cylinder
(or disk) about
cylinder axis
20. What is the initial rotational kinetic energy of the wheel?
21. What is the initial angular momentum of the wheel?
22. How much torque is required to stop the wheel in 6.0 seconds?
20. The initial rotational kinetic energy of the wheel is 442.125 J.
21. The initial angular momentum of the wheel is 29.55 kg m²/s.
22. The torque required to stop the wheel in 6.0 seconds is 4.925 Nm.
How to determine kinetic energy, angular momentum and torque?The initial rotational kinetic energy of the wheel can be calculated using the formula:
KE = (1/2) I ω²
where I = moment of inertia, ω = angular velocity. For a solid cylinder, the moment of inertia is (1/2)MR².
Plugging in the values:
I = (1/2)(5.6 kg)(0.28 m)² = 0.985 kg m²
ω = 30 rad/s
KE = (1/2) I ω² = (1/2)(0.985 kg m²)(30 rad/s)² = 442.125 J
Therefore, the initial rotational kinetic energy of the wheel is 442.125 J.
The initial angular momentum of the wheel can be calculated using the formula:
L = I ω
where I = moment of inertia, ω = angular velocity.
Plugging in the values:
I = 0.985 kg m²
ω = 30 rad/s
L = I ω = (0.985 kg m²)(30 rad/s) = 29.55 kg m²/s
Therefore, the initial angular momentum of the wheel is 29.55 kg m²/s.
The torque required to stop the wheel can be calculated using the formula:
τ = L/Δt
where L = angular momentum, and Δt = time interval.
Plugging in the values:
L = 29.55 kg m²/s
Δt = 6.0 s
τ = L/Δt = (29.55 kg m²/s)/6.0 s = 4.925 Nm
Therefore, the torque required to stop the wheel in 6.0 seconds is 4.925 Nm.
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