The distance covered by an accelerating object starting from rest is
D = (1/2 of acceleration) x (time²).
In this question,
D = 100 m
time = 3.8 s
100 = (1/2 acceleration) x (3.8s²)
100 = (1/2 a) x (14.44)
100/14.44 = 1/2 a
200/14.44 = a
a = 13.85 m/s²
22. On a day the wind is blowing toward the south at 3m/s, a runner jogs west at 4m/s. What is the velocity of the air relative to the runner?
The velocity of the air relative to the runner is 5 m/s.
What is the relative velocity?We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;
Velocity of the runner = 4m/s. due west
Velocity of the wind = 3m/s due south
The relative velocity is;
Vr = √(4)^2 + (3)^2
Vr = 5 m/s
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The cutoff frequency for a certain element is 1.22 x 10^15 Hz. What is its work function in eV?
The work function in eV for the given cutoff frequency is 5.05 eV.
What is cutoff frequency?The work function is related to the frequency as
Wo = h x fo
where, fo = cutoff frequency and h is the Planck's constant
Given is the cutoff frequency for a certain element is 1.22 x 10¹⁵ Hz
Wo = 6.626 x 10⁻³⁴ x 1.22 x 10¹⁵ Hz / 1.6 x 10⁻¹⁹
Wo = 5.05 eV
Thus, the work function is 5.05 eV
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A circular wheel of mass 50 kg and radius 200 mm is rotating at 300 r.p.m. Find its kinetic energy.
Please I need help fast. Assignment is due soon.
1) A ball is thrown from the top of a building with an initial speed of 8.0 m/s at an angle of 35° above the horizontal. The building is 18 m tall.
a) How long is the ball in the air?
b) How far from the building does the ball land?
c) What is its impact speed?
2) A 65-kg person driving a car hits the gas, accelerating the car at a rate of 3.9 m/s^2. Find the magnitude and direction of the force exerted by the seat on the person's body. Remember to include both the horizontal and the vertical components.
(a) The time spent in the air by the ball is 1.5 seconds.
(b) The horizontal distance of the ball is 9.83 m.
(c) The impact speed of the ball is 20.37 m/s.
(2) The magnitude of the force is 253.5 N in horizontal direction.
Time of motion of the ball
The time of motion of the ball is calculated as follows;
h = vt + ¹/₂gt²
18 = 8sin(35)t + (0.5)(9.8)t²
18 = 4.589t + 4.9t²
4.9t² + 4.589t - 18 = 0
solve the quadratic equation using formula method,
t = 1.5 s
Horizontal distance of the projectileX = vcosθ(t)
X = 8cos(35) x 1.5
X = 9.83 m
Impact speed of the projectilevyf = vyi + gt
vyf = 8sin(35) + 9.8(1.5)
vyf = 19.289 m/s
vxf = vxi
vxf = 8 x cos(35)
vxf = 6.55 m/s
Resultant speed = √(19.289² + 6.55²) = 20.37 m/s
Magnitude of the force exerted by the seat on the personF = ma
F = 65 x 3.9
F = 253.5 N
Since the motion is horizontal, the angle of the motion is zero.
Fx = 253.5cos(0) = 253.5 N
Fy = 253.5sin(0) = 0 N
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A 50 g bullet moving horizontally at a velocity of 400 m/s collides and embeds itself in a 25 kg block that rests on a rough horizontal surface and is attached to a spring with a spring constant of 180 N/m. The impact causes a compression of 12 cm in the spring. What is the coefficient of friction between the block and the surface?
The coefficient of friction between the block and the surface is determines as 0.1.
Final velocity of the block and the bulletApply the principle of conservation of linear momentum;
0.05(400) = v(0.05 + 25)
20 = 25.05v
v = 0.8 m/s
Coefficient of frictionApply the principle of conservation of energy
P.E - K.E = Ux
μmgx - ¹/₂mv² = ¹/₂kx²
μ(25.05)(9.8)(0.12) - 0.5(25.05)(0.8)² = 0.5(180)(0.12)
29.459μ - 8.016 = 10.8
29.459μ = 2.784
μ = 0.1
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Question 1 of 25
In the covalent compound CO₂, the Greek prefix used to represent the anion is
Answer:
di
Explanation:
The prefix is di. CO2 is carbon dioxide. di means 2.
Need help with this question!!
A beta particle or an electron is released during beta decay. The charges can be positive or negative. Co changes to Ni, Fe to Mn, Pb to Tl, and Pu to Am.
What is beta decay?Beta-decay is the radioactive decay that involves the release of the beta particle or the positron or the electron. The larger nucleus splits during nuclear fission and releases smaller nuclei.
The nuclear reactions are shown as:
⁶⁰Co₂₇ → ⁶⁰Ni₂₈ + ⁰e₋₁⁵⁶Fe₂₆ → ⁵⁶Mn₂₅ + ⁰e₋₁²¹⁰Pb₈₂ → ²¹⁰Tl₈₁ + ⁰e₋₁²⁴¹Pu₉₄ → ²⁴¹Am₉₅ + ⁰e₋₁Learn more about beta decay here:
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What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 600 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
[tex]q = -21 * 10^{-6} C[/tex]
What is Free-fall acceleration?The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.
As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,
mg =qE
[tex]1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\[/tex]
and its sign must be negative so that it will have upward electric force
so it is
[tex]q = -21 * 10^{-6} C[/tex]
The charge of a particle of mass is [tex]-21 * 10^{-6} C[/tex]
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a person walk at first constant speed 5 m/s then back alone the straight line from B to A at a constant speed 3 m/s a) what is his average speed over the entire trip b) what is his average velocity over the entire trip
Answer:
3.75 m/s
Explanation:
Comment
Let's deal with the second question first. The average velocity is 0 because the displacement (distance between the starting point and ending point) is 0.
The first question is a little harder. You don't seem to have enough information. When that happens, you can just make things up. Now there's an interesting solution. You could do it with algebra, but it is easier to see with numbers.
Givens
d = 150 m. between A and Br1 = rate from A to B = 5 m/sr2 = rate from B to A = 3 m/stime A to B = t1time B to A = t2Formulas
t1 = d / r1
t2 = d/r2
average rate = total distance / total time
Solution
t1 = 150 m / 5 m/s = 30 seconds
t2 = 150 m / 3 m/s = 50 seconds
Total distance = 150 m + 150 m = 300 m
Total time = 30 seconds + 50 seconds = 80 seconds
Average speed = 300 m / 80 s
Answer
Average speed = 3.75 m/s
what's the difference between coplanar forces and resultant forces?
Answer:
coplanar When all forces are acting in the same
resultunt the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body via vector addition
If have a custom made telescope with these specifications then what's the focal length of my telescope? (refractor telescope only please)
_Specifications:_
*Objective diameter*: 90mm
*Distance between lenses*: 230mm
*Eyepiece focal length*: 4mm
*Eyepiece size*: 31.7mm
Please respond fast. Also please give a simplified answer for a brainliest answer tag. Thanks!
Answer:
i don't know sorry
i know a good website that may help you
it's called https://skyandtelescope.org/observing/telescope-calculator/
Explanation:
Two students have fitted their scooters with the same engine. Student A and his
scooter have a combined mass of 127.5 kg and a maximum acceleration of
2.40 m/s². Student B has a maximum acceleration of 1.70 m/s on her scooter.
Show that the combined mass of student B and her scooter is 120 kg
You may assume that the frictional forces acting on each scooter are negligible.
The force exerted by student A with his scooter is 306 N and that of student B is 204 N.
Force applied by each studentThe force exerted by each student is calculated from Newton's second law of motion.
F = ma
where;
m is mass a is accelerationF(A) = 127.5 x 2.4
F(A) = 306 N
F(B) = 120 x 1.7
F(B) = 204 N
Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.
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What is the acceleration of Karla's I Phone being thrown from Mr. Higley's classroom at 0m/s if it hits the wall 1.2 seconds later going 35 m/s?
The acceleration of Karla's I IPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²
What is acceleration?The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.
The given data in the problem is given by ;
u is the initial speed = 0 m/sec
v is the final speed= 35 m/sec
t is the time interval= 1.2 second
a is the acceleration=? m/sec²
The formula for acceleration is;
[tex]\rm a=\frac{v-u}{t} \\\\ a= \frac{35-0}{1.2} \\\\ a= 29.16 \ m./s^2[/tex]
Hence, the acceleration of Karla's iPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²
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Which two of the following statements best describe the similarity between elements and compounds? 1. Elements and compounds are listed in the periodic table. 2. Elements and compounds are both pure substances. 3. Elements and compounds are both made of different kinds of atoms. 4. Elements and compounds cannot be broken down by physical changes.
As a warehouse worker pushes a crate across a concrete floor, the force he
applies is not perfectly horizontal, as shown in the image below. If the
coefficient of kinetic friction between the crate and concrete floor is 0.5, what
is the net force on the crate?
A. 136 N
B. 99 N
C. 73 N
D. 112 N
The net force on the crate will be 99 N.Option B is correct.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). it is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and acceleration.Mathematically in the different components and balancing the equation gets.Components in the x-direction.
When all the forces get resolved the y-direction of forces are;
300 sin 10° +N = 445
N = 445-295.4
N = 392.9 N
The normal force is 392.9 N.
The net force on the crate is found by resolving the force in the x-direction;
F = 300 cos 10° - μN
F= 295.44-196.45
F= 98.99
F=99
Hence, the net force on the crate will be 99 N.Option B is correct.
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What will be the temperature reading in Fahrenheit scale whe eading is 310 k?
The temperature reading in Fahrenheit scale when reading 310 K is 98.6F
What is temperature?
Temperature is the degree of hotness or coldness of the object.
Given is a temperature in Kelvin scale = 310 K
Temperature on centigrade scale = 310 -273 = 37°C
Temperature conversion from Centigrade to Fahrenheit is
C/5 = F -32/9
Plug the value, we get
37/5 = F-32/9
66.6 = F -32
F = 32+66.6 =
F = 98.6
Thus, the temperature of 310 K in Fahrenheit is 98.6F.
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An upward force is applied to a 6.0–kilogram box. This force displaces the box upward by 10.00 meters. What is the work done by the force on the box?
The work done by the force on the box 588 Joule.
What is Potential energy ?It is the energy that is stored in an object due to its position relative to some zero position.
According to the question,
Given mass of the box (m) = 6.0 kg
Given displacement of the box = 10.00 m
Hence , the height of the box (h) = 10.00 m
Acceleration due to gravity (g) = 9.8 m/s²
Since ,
The upward force was applied to the box, the box moves up to the height.
Therefore, the work done by the ball is the potential energy gained by the ball.
By using the formula of potential energy
P.E. = mgh
= 6 × 9.8 × 10 Joule = 588 Joule
As we know .
Work done (W) = Potential energy (P.E.)
Work done (W) = 588 Joule
Thus, 588 Joule is the work done by the force on the box.
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3
012 10.0 points
A car with mass 1060 kg crashes into a wall.
The car's velocity immediately before the col-
lision was 14.6 m/s and the bumper is com-
pressed like a spring with spring constant
1.14 × 107 N/m.
What is the maximum deformation of the
bumper for this collision?
Answer in units of m.
013 10
Hi there!
The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.
Initially, we only have kinetic energy:
[tex]KE = \frac{1}{2}mv^2[/tex]
KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)
Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:
[tex]U_s = \frac{1}{2}kx^2[/tex]
k = Spring Constant (1.14 × 10⁷ N/m)
x = compressed distance of bumper (? m)
Since energy is conserved:
[tex]E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2[/tex]
We can simplify and solve for 'x'.
[tex]mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}[/tex]
Plug in the givens and solve.
[tex]x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}[/tex]
find the velocity and acceleration if x=t³ +2t²+4 at a time 2s
Answer:
[tex]\textsf {v = 20 m/s}[/tex]
[tex]\mathsf {a = 16 m/s^2}[/tex]
Explanation:
[tex]\textsf {To find velocity, let's differentatiate x :}[/tex] [tex]\implies \mathsf {v = \frac{dx}{dt} }[/tex]
⇒ [tex]\mathsf {v = \frac{d}{dx}(t^{3}+2t^{2}+4) }[/tex]
⇒ [tex]\mathsf {v = 3t^{2} +4t +0 }[/tex] [tex]\textsf {(By applying law of differentiation)}[/tex]
[tex]\textsf {Substitute t = 2 :}[/tex]
⇒ [tex]\mathsf {v = 3(2)^{2} + 4(2) }[/tex]
⇒ [tex]\textsf {v = 12 + 8}[/tex]
⇒ [tex]\textsf {v = 20 m/s}[/tex]
[tex]\textsf {To find acceleration, let's differentatiate v :}[/tex] [tex]\implies \mathsf {a = \frac{dv}{dt} }[/tex]
⇒ [tex]\mathsf {a = \frac{d}{dx}(3t^{2}+4t) }[/tex]
⇒ [tex]\textsf {a = 6t + 4}[/tex]
[tex]\textsf {Substitute t = 2 :}[/tex]
⇒ [tex]\textsf {a = 6(2) + 4}[/tex]
⇒ [tex]\textsf {a = 12 + 4}[/tex]
⇒ [tex]\mathsf {a = 16 m/s^2}[/tex]
imagine your lifting a toy car above your head and then holding it stationary above your head. While it is stationary what forces do you think act on the toy car
Gravity and normal forces act on the toy car.
What is force?A push or pull that an object experiences as a result of interacting with another item are known as a force. Every time two items interact, a force is exerted on each of the objects. The force is no longer felt by the two objects when the interaction ends. Only when there is interaction do forces actually exist.Both contact forces and forces coming from the action at a distance can be used to categorize all forces (interactions) between objects.Frictional forces, tensional forces, normal forces, air resistance forces, and applied forces are a few examples of contact forces.Gravitational forces are examples of action-at-a-distance forces.The standard metric unit known as the Newton is used to quantify force. "N" stands for Newton in abbreviation.To learn more about force refer to:
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A force of 5.25 newtons acts on an object of a mass 25.5 kilograms. How far is the object from the center of Earth? (The value of G is 6.673 × 10-11 newton meter2/kilogram2. The mass of Earth is 5.98 × 1024 kilograms.)
Answer:
The force between two objects is calculated through the equation,
F = Gm₁m₂/d²
where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is the universal gravitation constant.
In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,
2.5F = Gm₁m₂ / (D²)
D = 0.623d
Subsituting the known value of d,
D = 0.623(6.9 x 10^8) = 4.298 x 10^8 m
It just so happens that regardless of the material, when objects are heated up they will start to glow and change colors at near identical temperatures. The plot that you see is called a blackbody spectrum. This plot tells us the intensity or the “amount” of light that an object will emit at different wavelengths (or “colors”). The visible wavelengths are marked by their colors on the plot. To the right of the visible band is lower energy infrared light. To the left of this band is higher energy ultraviolet (UV) light.
Click the + button that is to the left of the intensity scale (far left side of the screen) such that the top of the scale is at 1x10-3 (in the picture above, the top of the scale says 100).
Now use the temperature slider to the right, and take the temperature all the way down to 300 Kelvin (80 Fahrenheit).
Now slowly begin to raise the temperature. At approximately what temperature would a heated material (metal, wood, etc.) begin to give off visible light at a deep red color?
Answer: the plot that you see is called a blackbody spectrum. This plot tells us the intensity or the amount of light that an object will emit at different wavelenths.
Explanation: i hope that this helps
Radioactive decay can involve electrons
True or False
Answer:
i believe the answer is true
question is below............
The total distance traveled by the gardener is 8 meters. the magnitude of total displacement is approximately 7 meters.
Total distance covered by gardener = 2 + 1 +2 + 3= 8 meters
For the magnitude of displacement, we will take the shortest path for the gardener movement.
for displacement-
[tex]=\sqrt{( 0-4^{2}) +( 6-0^{2} )}\\=\sqrt{52} \\=7[/tex](approximately)
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An m = 6.11 g bullet is fired into a 365 g block that is initially at rest at the edge of a table of h = 1.08 m height as shown in the figure.
1. The bullet remains in the block, and after the impact the block lands d = 1.84 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?
2. What is the initial horizontal velocity of the block as it flies off the table?
3. Determine the initial speed of the bullet.
(a) The time taken the block to reach the ground once it flies off the edge of the table is 0.47 s.
(b) The initial horizontal velocity of the block as it flies off the table is 3.91 m/s.
(c) The initial speed of the bullet is 237.5 m/s.
Time of motion of the blockThe time of motion of the block is calculated as follows;
h = ¹/₂gt²
where;
h is height of the tablet is time of motiont = √(2h/g)
t = √( (2 x 1.08) / 9.8)
t = 0.47 s
Initial horizontal velocity of the blockx = vt
v = x/t
v = (1.84)/(0.47)
v = 3.91 m/s
Initial speed of the bulletm1u1 + m2u2 = v(m1 + m2)
where;
m1 is mass of bulletm2 is mass of blocku1 is initial velocity of the bulletu2 is the initial velocity of the blockv is initial horizontal velocity of the block as it flies off the table0.00611(u1) + 0.365(0) = 3.91(0.00611 + 0.365)
0.00611u1 = 1.451
u1 = 1.451/0.00611
u1 = 237.5 m/s
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A surfer on top of a wave takes 10 seconds to travel a distance of 40 meters. What is the speed of the wave the surfer is riding on?
Select one:
a. 0.25 m/s
b. 4 m/s
c. 40 m/s
d. 400 m/s
Answer:
B. 4 m/sExplanation:
The formula for Speed is distance per time.
Therefore Speed = Distance / TimeData
D = 40 m
T = 10 sec
S = ?
Speed = 40m /10sec
To find the speed , divide the distance by the time.
Note : the unit for Speed is meter /second
Speed = 4 m/sTherefore the speed of the wave, the surfer is riding on is 4 m/s.
i need help with lab
(a) The equilibrant C for force of vector A and B is 3.43 N.
(b) The equilibrant C for fx of vector A and B is 2.1 N.
(c) The equilibrant C, for fy of vector A and B is 2.12 N.
What is equilibrant force?An equilibrant force is a single force that will bring other bodies into equilibrium.
From configuration 1:Vector A: mass = 0.2 kg, θ = 20⁰
Vector B: mass = 0.15 kg, θ = 80⁰
Fx = mg cosθ
Fy = mg sinθ
where;
m is mass g is acceleration due to gravityVector AForce of A due to its weight
F(A) = mg
F(A) = 0.2 x 9.8 = 1.96 N
Fx = (0.2 x 9.8) cos(20) = 1.84 N
Fy = (0.2 x 9.8) sin(20) = 0.67 N
Resultant forceR = √(0.67² + 1.84²)
R = 1.96 N
Vector BForce of B due to its weight
F(B) = mg
F(B) = 0.15 x 9.8
F(B) = 1.47 N
Fx = (0.15 x 9.8) cos(80) = 0.26 N
Fy = (0.15 x 9.8) sin(80) = 1.45 N
Resultant forceR = √(0.26² + 1.45²)
R= 1.47 N
Equilibrant C of vector A and BEquilibrant force:
Force, C = 1.96 N + 1.47 N
Force, C = 3.43 N
Equilibrant FX:
Fx, C = Fx(A) + Fx(B)
Fx, C = 1.84 N + 0.26 N = 2.1 N
Equilibrant FY:
Fy, C = Fy(A) + Fy(B)
Fy, C =0.67 N + 1.45 N = 2.12 N
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Which is the correct unit for electrical power?
Answer:
As electrical power is measured in Watts (W), therefore it must be also be measured in Joules per Second. So we can correctly say that: 1 watt = 1 joule per second (J/s).
Explanation:
loade Brack Figure below show a circuit diagram of a device for controlling the temperature in a room. Explain how the Iron Heating element S loade Brack Figure below show a circuit diagram of a device for controlling the temperature in a room . Explain how the Iron Heating element S
Answer:
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if you were to draw a 3rd harmonic of a tube open at both ends, what would you draw at the ends of the tube?
A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.
In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.
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