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 sports car can move 100.0 m in the first 3.8 s of constant acceleration. What is the car's acceleration?
m/s²
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
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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Tennis balls experience a large drag force. A tennis ball is hit so that it goes up and then comes back straight down.
A tennis ball is hit by a large force so that it goes up into the air and then it comes back straight down because of gravity.
How object move upward and downward?We know that objects move upward due to application of force on it while on the other hand, object comes to the ground because of the attraction of earth which we called gravity.
So we can conclude that a tennis ball is hit by a large force so that it goes up into the air and then it comes back straight down because of gravity.
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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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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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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
Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 5.5 cm per year.
(a) What distance does it move in 51 seconds at this speed?
m
(b) What is its speed in miles per million years?
mi/My
The distance and speed are as follows:
Distance = 8.89 * 10⁻⁶ cmSpeed = 34.2 miles per million years.What is speed of a body?Speed of a body is the ratio of the distance covered by a body and the time it takes to cover that distance,
Speed = distance/timeThe number of seconds in a year = 365.25 * 24 * 3600 = 31557600 second
a) Distance covered in 51 seconds = 5.5/31557600 * 51
Distance = 8.89 * 10⁻⁶ cm
b) 5.5 cm = 3.4175 * 10⁻⁵ miles
Sped in miles per million years = 3.42 * 10⁻⁵ miles/1 * 10⁻⁶ million years
Speed = 34.2 miles per million years.
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Entropy is how quickly things get messy.
O A. True
OB. False
Answer : False
Answer:
false
Explanation:
it cant defined the messy and clean states
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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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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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:
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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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.
Radioactive decay can involve electrons
True or False
Answer:
i believe the answer is true
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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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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A skater starts from rest v1=0, from the top of an h=6m hill and travels d=20m until it gets to the bottom of the hill where its speed is v2=1.2m/s. If the skater has m=50kg, what is the friction force?
The friction force is 1.5N
skater's initial velocity v1=0height of the hill=20mskater's final velocity=1.2m/sMass of the skater=50kg
Now, using the third equation of motion-
[tex]v^{2}= u^{2} +2as[/tex]
v=final velocity
u=initial velocity
a=acceleration
s=distance
substitute the values in the above equation
1.2=0+ 2a×20
a=1.2/40
a=0.03
friction force=mass ×acceleration
=50×0.03
=1.5N
hence, The friction force is 1.5N
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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]
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]
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
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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Which sentence uses the subjunctive mood correctly?
A. Americans take fewer vacations than people in many other
nations.
OB. It's best that Marci learns how to do her own laundry.
C. It is important that she apologize to her sister for what she said.
D. Be sure to put your dirty clothes in the hamper, not next to it.
The sentence that uses the subjunctive mood correctly is as follows:
Be sure to put your dirty clothes in the hamper, not next to it.Thus, the correct option is D.
What is Subjunctive mood?Subjunctive mood may be defined as statements or actions that are in the probable situation of doubt and suspicion.
Except for the sentence fourth, the rest all indicates the factual concept with the appropriate condition.
Therefore, the correct option for this question is D.
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Answer:
C. it is important that she apologize to her sister for what she said
Explanation:
test approved
A circular wheel of mass 50 kg and radius 200 mm is rotating at 300 r.p.m. Find its kinetic energy.
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 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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Ivan exerts 20 N of force to lift a basket 2 m. He then carries the basket 5 m across a room to place it on a shelf.
Ivan does work when he
.
The amount of work he does is
The amount of the work done will be 100 J. Work done is denoted by W.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.
The given data in the problem is;
F is the force applied = 20 N
d is the displacement = 5.0 m
The work done is found as;
W = F × d
W = 20 N × 5
W= 100 J
Hence, the amount of work done will be 100 J.
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Answer:
lifts the basket, 40J
Explanation:
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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Two masses, m and 2m, approach each along a path at right angles to each other. After collision, they stick together and move of at 2m/s at angle 37⁰ to the original direction of the mass m. What where the initial speeds of the two particles?
The initial speed of the object with mass "m" is 4.79 m/s and the initial speed of the object with mass "2m" is 1.81 m/s.
Initial speed of the two massesThe initial speed of the two masses is calculated by applying the principle of conservation of linear momentum as follows;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
let m be in horizontal directionlet 2m be in vertical directionangle between the two masses = 90 degreesin y - direction[tex]m_1u_1sin(\theta_1) + m_2u_2sin(\theta_2) = (m_1 + m_2)vsin(\theta_3)\\\\mu_1sin(0) + 2mu_2sin(90) = (m + 2m)(2)sin(37)\\\\2mu_2 = 3.61m\\\\u_2 = \frac{3.61 m}{2m} \\\\u_2 = 1.81 \ m/s[/tex]
in x - direction[tex]m_1u_1sin(\theta_1) + m_2u_2sin(\theta_2) = (m_1 + m_2)vsin(\theta_3)\\\\mu_1cos(0) + 2mu_2cos(90) = (m + 2m)(2)cos(37)\\\\mu_1 = 4.79 m\\\\u_1 = 4.79 \ m/s[/tex]
Thus, the initial speed of the object with mass "m" is 4.79 m/s and the initial speed of the object with mass "2m" is 1.81 m/s.
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