Given that the slope of the ramp is L = 8 m
The height of the ramp is h = 2 m
We need to calculate the mechanical advantage of the ramp.
The mechanical advantage of the ramp can be calculated by the formula
[tex]\text{Mechanical advantage =}\frac{L}{h}[/tex]Substituting the values, the mechanical advantage will be
[tex]\begin{gathered} \text{Mechanical advantage =}\frac{8}{2} \\ =4\text{ } \end{gathered}[/tex]Thus, the correct option is A. 4
A wheel of radius 30.0 cm is rotating at a rate of 2.30 revolutions every 0.0810 s.Through what angle does the wheel rotate in 1.00 s?What is the linear speed of a point on the wheel’s rim?What is the wheel’s frequency of rotation?
Given data
*The given radius of the wheel is r = 30.0 cm = 0.30 m
*Rate of rotation is 2.30 revolutions every 0.0810 s
The angle of the wheel rotates in one second is calculated as
[tex]\begin{gathered} \theta=(2.30\times2\pi)\times\frac{1}{0.0810} \\ =178.32\text{ radian} \end{gathered}[/tex]Hence, the angle of the wheel rotating in one second is 178.32 radian
The formula for the linear speed of a point on the wheel's rim is given as
[tex]v=r\omega[/tex]Substitute the known values in the above expression as
[tex]undefined[/tex]A coil of 17.771 H carries a current of 11.121 A. Compute the energy stored .
Given
The inductance is H=17.771 H
Current is , I=11.121A
To find
The energy stored
Explanation
The energy stored is given by
[tex]E=\frac{1}{2}LI^2[/tex]Putting the values,
[tex]\begin{gathered} E=\frac{1}{2}\times17.771\times(11.121)^2 \\ \Rightarrow E=1098.92J \end{gathered}[/tex]Conclusion
The energy stored is 1098.92J
An athlete starts at point A and runs at a constant speed of around a circular track 100 m in diameter
, as shown in Fig. P3.40 below. Find the x and y-components of this runner’s average velocity and average acceleration between points
(a) A and B, (b) A and C, (c) C and D, and (d) A and A (a full lap). (e) Calculate the magnitude of the runner’s average velocity between A and B. Is his average speed equal to the magnitude of his average velocity? Why or why not? (f) How can his velocity be changing if he is running at constant speed?
a ) The x and y-components of average velocity and average acceleration between points A and B are 3.8 m/s, 3.8 m/s and 0.46 m/s², - 0.46 m/s²
e ) The magnitude of the runner’s average velocity between A and B is
t = 2 π r / v
t = 2 * 3.14 * 50 / 6
t = 52.4 s for full lap
t per quarter = 52.4 / 4 = 13.1 s
v = Δx / Δt
a = Δv / Δt
a ) From A to B,
vx = ( 0 - ( - 50 ) ) / 13.1
vx = 3.8 m / s
vy = ( 50 - 0 ) / 13.1
vy = 3.8 m / s
ax = ( 6 - 0 ) / 13.1
ax = 0.46 m / s²
ay = ( 0 - 6 ) / 13.1
ay = - 0.46 m / s²
b ) From A to C,
t = 52.4 / 2
t = 26.2 s
vx = ( 50 - ( - 50 ) ) / 26.2
vx = 3.8 m / s
vy = 0
ax = 0
ay = ( - 6 - 6 ) / 26.2
ay = - 0.46 m / s²
c ) From C to D,
t = 13.1 s
vx = ( 0 - 50 ) / 13.1
vx = - 3.8 m / s
vy = ( - 50 - 0 ) / 13.1
vy = - 3.8 m / s
ax = ( - 6 - 0 ) / 13.1
ax = - 0.46 m / s²
ay = ( 0 - ( - 6 ) ) / 13.1
ay = 0.46 m / s²
d ) From A to A,
Since the starting and ending points are exactly the same, there is no displacement. So the average velocity will be zero. Due to no change in velocity, there will be no acceleration
e ) From A to B,
v = √ vx² + vy²
v = √ 3.8² + 3.8²
v = 5.4 m / s
Displacement is the shortest distance between two points. So it will basically be a straight line. But the athlete runs in a circular motion. So distance will be larger than the displacement. So speed will be higher than velocity.
s = 6 m / s
v = 5.4 m / s
s > v
f ) At constant speed in a circular motion, only the magnitude is constant. Its direction keeps changing. So velocity cannot be constant in a circular motion.
Therefore,
a ) vx = 3.8 m / s, vy = 3.8 m / s ; ax = 0.46 m / s², ay = 0.46 m / s²
b ) vx = 3.8 m / s, vy = 0 ; ax = 0, ay = - 0.46 m / s²
c ) vx = - 3.8 m / s, vy = - 3.8 m / s ; ax = - 0.46 m / s², ay = 0.46 m / s²
d ) vx = 0, vy = 0 ; ax = 0, ay = 0
e ) v = 5.4 m / s
f ) Due to change in direction.
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A comet is known to be traveling at 236 km per second. It is observed to give light with a frequency of 641 x 1014 Hz. What frequency of light would the come be emitting if it were motionless? Express your answer to the nearest hundred trillion Hz.
ANSWER
6.4 x 10¹⁶ Hz
EXPLANATION
Given:
• The speed of the source, vs = 236 km/s
,• The speed of light, c = 300,000 km/s
,• The observed frequency of the light, fo = 641 x 10¹⁴ Hz
Find:
• The frequency of the source, fs
To solve this problem, we have to apply the Doppler Effect formula for light,
[tex]f_o=f_s\sqrt{\frac{1-v_s/c}{1+v_s/c}}[/tex]Solving for fs,
[tex]f_s=f_o\sqrt{\frac{1+v_s/c}{1-v_s/c}}[/tex]Replace the known values and solve,
[tex]f_s=641\cdot10^{14}Hz\cdot\sqrt{\frac{1+(236km/s)/(300,000km/s)}{1-(236km/s)/(300,000km/s)}}\approx6.4\cdot10^{16}Hz[/tex]Hence, the emitted frequency of the comet's light is 6.4 x 10¹⁶ Hz.
A company is constructing a building for a recordings tudio. Which of the following would be the LEAST HELPFUL in reducing road noise in the studio?
The lest helpful solution is reduncing the number of trees on the property. Therefore, the answer is A.
This comes from the fact that all the other option will better absorb sound waves on the building itself.
A battery does 1.922 J of work to transfer 0.089 C of charge from the negative to the positive terminal. What is the emf of this battery?
We can find the EMF as follows:
[tex]\begin{gathered} EMF=\frac{W}{q} \\ where: \\ W=1.922J \\ q=0.089C \\ so: \\ EMF=\frac{1.922}{0.089} \\ EMF=21.5955V \end{gathered}[/tex]Answer:
21.5955V
A satellite is in orbit around Earth. How does the force exerted by Earth on the satellite compare to the force exerted by the satellite on Earth? Note: >> means much greater than; << means much less than.
The formula for the gravitational force between two bodies is given as
[tex]F=\frac{GmM}{r^2}[/tex]*Here m is the mass of the first object
*Here M is the mass of the second object
*Here r is the distance between the two objects
*Here G is the gravitational force
The force exerted by the earth on the satellite is equal to the force exerted by the satellite on the earth. Hence, the correct option is (b)
Three liquids are at temperatures of 10 ◦C, 22◦C, and 29◦C, respectively. Equal masses of the first two liquids are mixed, and the equi- librium temperature is 12◦C. Equal masses of the second and third are then mixed, and theequilibrium temperature is 25.9◦C.Find the equilibrium temperature when equal masses of the first and third are mixed.Answer in units of ◦C.
The belief that for every particle there is a force particle is referred to in modern physics as which of the following?Select one:a.unification theoryb.string theoryc.supersymmetryd.quantum tunneling
We will have that the one would be supersymmetry.
I have a homework problem that I don’t even know where to start or what formulas to use
We will have the following:
a) The velocity after 5 seconds will be:
[tex]v_f=v_o+at[/tex][tex]v=2.79m/s+(9.8m/s^2)(5s)\Rightarrow v=51.79m/s[/tex]So, the velocity after 5 seconds will be 51.79m/s.
b) We will have that the distance below the helicopter will be:
[tex]d=\frac{1}{2}(v_0+v_f)t[/tex][tex]d=\frac{1}{2}(2.79m/s+51.79m/s)(5s)\Rightarrow d=136.45m[/tex]So, it will be 136.45 m below the helicopter.
c) We will have that the velocity and distance given that the helicopter is moving constantly at 2.79m/s will be:
[tex]v=-2.79m/s+(9.8m/s^2)(5s)^2\Rightarrow v=46.21m/s[/tex]So, the velocity would be 46.21 m/s.
[tex]d=\frac{1}{2}(-2.79m/s+46.21m/s)(5s)\Rightarrow d=108.55m[/tex]So, the distance would be 108.55 m below the helicopter.
Which statement given BEST describes what happens to light as it passes from air into a piece of glass?a) The speed increases, its wavelength becomes longer, and its frequency decreases.b)The speed increases, its wavelength becomes longer, and its frequency remains the same.c) The speed decreases, its wavelength becomes shorter, and its frequency remains the same.d) The speed decreases, its wavelength becomes shorter, and its frequency increases.
Given that the light passes from air to glass, that means it enters to denser medium from rarer medium.
Here, air is the rarer medium and glass is the denser medium.
Speed decreases when light enters from rarer to densr medium , while the frequency remains the same and its wavelength becomes shorter.
Thus, option C is correct.
Block A hangs from a light string that passes over a light pulley and is attached to block B, which is on a level horizontal frictionless table as shown above. Students are to determine the mass of block B from the motion of the two-block system after it is released from rest. They plan to measure the time block A takes to reach the floor. The students must also take which of the following measurements to determine the mass of block B?A. Only the mass of block A.B. Only the mass of Block A falls to reach the floorC. Only the mass of block A and the distance block A falls to reach the floor D. The mass of block A, the distance block A falls to reach the floor, and the radius of the pulleyUsing symbols like mA, mB, g, and a, wrote an equation that expresses Newton’s Second Law on the entire system as a whole. (Hint: Fnet = m•a) Which letter in the equation you just wrote would need to be measured before you could solve for mB? Assume you already know the value of g. One of the letters you should have written was a. The students are going to measure the time it takes for block A to reach the floor. What other quantity needs to be measured to find the acceleration? What equation would need to be used to find the acceleration?
Required: the mass of the block B.
Explanation:
we assume that mass of the block A is
[tex]m_A[/tex]and mass of the block B is
[tex]m_B[/tex]now, look at the free body diagram
from the above diagram, we can apply newton's law. we assume that both the block moves with the same acceleration a.
for block A
[tex]m_Ag-T=m_Aa.....(1)[/tex]for block B
[tex]T=m_Ba[/tex]from equations 1 and 2
we can write
[tex]\begin{gathered} m_Ag-m_Ba=m_Aa \\ a=\frac{m_Ag}{m_A+m_B}........(2) \end{gathered}[/tex]this is the acceleration of the whole system.
By the above equation, we can calculate the acceleration of the both the blocks.
we are interested in determining the mass of block mB.
we assume that block mA moves with above acceleration.
we know that
[tex]h=ut+\frac{1}{2}at^2......(3)[/tex]if we know the height of the block A and measure the time to reach the block A to the ground.
by the equation 3 we can calculate the acceleration.
[tex]a=\frac{2h}{t^2}......(4)[/tex]from the equation 2 and 4, we can write
[tex]\begin{gathered} \frac{2h}{t^2}=\frac{m_{A}g}{m_{A}+m_{B}} \\ m_A+m_B=\frac{t^2m_Ag}{2h} \\ m_B=\frac{t^{2}m_{A}g}{2h}-m_A......(5) \end{gathered}[/tex]by the above equation, we can easily can calculate the mass of the block B.
(a) part
only the mass of the block A and the distance block A falls to reach the floor is correct answer.
As we can see from the above equation.
I have this homework i need help with ?Part (a)
Given that the mass of train is
[tex]m=\text{ }8.12\times10^6\text{ kg}[/tex]The deceleration of the train is
[tex]\begin{gathered} a\text{ = -104 km/h} \\ =-104\times\frac{5}{18} \\ =\text{ -28.88 m/s} \end{gathered}[/tex]The braking force can be calculated by the formula
[tex]F=ma[/tex]Substituting the values, the braking force will be
[tex]\begin{gathered} F=8.12\times10^6\times(-28.88) \\ =2.348\times10^8\text{ N} \end{gathered}[/tex]Thus, the braking force is 2.348 x 10^8 N.
If an alarm clock is ringing what is the medium for the sound waves
Sound energy travels through air in waves. When an alamr clock rings, nearby air molecules vibrate.
Identify the Action and Reaction Forces
A pair of forces that will always move in conflict with one another is referred to as an action-reaction force.
What is newton's third law?The third law of Newton asserts because when bodies interact, they exert forces that are equal in size and directed in the reverse way. Another name for the newton's third is the law of action and reaction. This equation is crucial for understanding issues with static balance, in which all forces are all in balance, but it also holds true for bodies moving at a steady or accelerated speed.
Hammer hitting a nail : In this, the hitting by the hammer is the action and the nail is going into the surface is the reaction force.
Book placed on the table : In this book is applying gravitational force that is action force and the reaction force is applied by the table.
Hitting a tennis ball : In this, hitting the ball is the action and the ball is going in the direction of hitting is the reaction force.
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Part 1) If a force of magnitude 125 N is exerted horizontally as shown, find the force exerted by the hammer claws on the nail. (Assume that the force the hammer exerts on the nail is parallel to the nail).Answer in units of N. Part 2) Find the force exerted by the surface on the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail. Answer in units of N.
Part 1)
Recall, for a body to remain in rotational and translational equilibrium, the net external torque and force acting on a system must be zero. The formula for calculating torque is
Torque = Force x moment arm
Let P be the force exerted by the hammer on the nail. For rotational equilibrium, the torque about the point of contact would be equal. Thus,
For the nail,
distance = 5.86 cm = 5.86/100 = 0.0586 m
moment arm = 0.0586mSin26 = 0.0257
Torque = P x 0.0257 = 0.0257P
For the handle,
Force = 125
perpendicular distance = 28 cm = 28/100 = 0.28
Torque = 125 x 0.28 = 35
By equation both torques,
0.0257P = 35
P = 35/0.0257
P = 1361.87 N
the force exerted by the hammer claws on the nail is 1361.87 N
Part 2)
The force exerted by the surface on the point of contact is the normal force, N
It would be equal to the total downward force. Since the claw is inclined,
Normal force = Downward force x Sinθ
Downward force = P = 1361.87
θ = 26
N = 1361.87Sin26
N = 597 N
the force exerted by the surface on the point of contact with the hammer head is 597 N
A 6.00 kg pendulum bob is placed on a 10.0 m long string and pulled back 5.00°. What is the period of the pendulum when it is released?3.84 s8.03 s6.35 s1.27 s
As we know that time period for a pendulum is,
[tex]\begin{gathered} T=2\pi\sqrt[\placeholder{⬚}]{\frac{l}{g}} \\ Here, \\ l=10m \\ g=9.8\text{ m/s}^2 \\ So, \\ T=2\times3.14\sqrt[\placeholder{⬚}]{\frac{10}{9.8}}=6.343s \\ \end{gathered}[/tex]So 3rd option is correct option.
A 98 N person is standing on a board, 1 m from the end. The board is balanced on a point that is 2m from the same end. The board is 49 N. How long is the board overall?
ANSWER:
8 m
STEP-BY-STEP EXPLANATION:
To calculate the length is the board overall we must apply the center of mass formula, which is as follows:
[tex]x_{cm}=\frac{m_{\text{board}}\cdot x_{\text{noard}}+m_p\cdot x_p}{m_{\text{board}}+m_p}[/tex]The mass of the person and that of the board, we calculate it as follows:
[tex]\begin{gathered} F_{\text{p}}=m_{\text{p}}\cdot g_{} \\ m_{\text{p}}=\frac{F_{\text{p}}}{g}=\frac{98}{9.8}=10kg \\ F_{\text{board}}=m_{\text{board}}\cdot g_{} \\ m_{\text{board}}=\frac{F_{\text{board}}}{g}=\frac{49}{9.8}=5kg \end{gathered}[/tex]Replacing:
[tex]\begin{gathered} -2=\frac{5\cdot x_{\text{board}}+10\cdot(-1)}{5+10} \\ (-2)(15)+10=5\cdot x_{\text{board}} \\ x_{\text{board}}=\frac{-20}{5} \\ x_{\text{board}}=-4\text{ m} \end{gathered}[/tex]Since the CM of the board is only 4 m from the edge of the board, and the MC of the board is at its center, the board is 8 m long.
what is the weight of a 500 g package of spaghetti noodles ? (A) in N (B) in oz
Given:
Weight of the spaghetti noddles = 500 grams
Let's find the weight of the noodles in Newtons(N) and Ounce(oz).
First convert 500g to kg:
[tex]500g=\frac{500}{1000}=\frac{1}{2}kg[/tex](A) in N
Where 1 kg = 9.8 N
Thus, we have:
[tex]\frac{1}{2}kg=\frac{1}{2}\ast9.8=4.9N[/tex]The weight of a 500 g package of spaghetti noodles in Newton is 4.9 N
(B) in oz
Where:
1 gram = 0.035274 ounce
Thus, we have:
[tex]500g=500\ast0.035274=17.637\text{ oz}[/tex]The weight of a 500g package of spaghetti noodles in Ounce is 17.637 oz
ANSWER:
(A) 4.9 N
(B) 17.637 oz
The carnival ride, the Round Up, has a radius of 1 m and rotates once each .9 s. What coefficient of static friction is required to keep the riders from slipping?
The coefficient of static friction is required to keep the riders from slipping is 0.05.
What is coefficient of static friction?The coefficient of static friction is the ratio of the maximum static friction force (F) between the surfaces in contact before movement commences to the normal (N) force.
The coefficient of static friction is required to keep the riders from slipping is calculated as follows;
static friction force = centripetal force
μmg = ma
μg = a
μg = ω²r
μ = ω²r /g
where;
ω is the angular speed of the carnival rider is the radius of the pathg is acceleration due to gravityThe given parameters;
angular speed of the carnival ride, ω = 1 rev/9s = 2π rad/9s = 0.698 rad/s
μ = (0.698² x 1) / (9.8)
μ = 0.05
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I have a question of a test that I already took and just want to know why my answer was incorrect, so when I spoke to the prevuios tutor and gave hime the question wuth the possible answer he asked me if it was for a graded test , and I said that it was a graded test and he said to report me , my question is a tutor can't help with an answer of a test that I took and my answer was incorrect, and I just want to understand why is it incorrect
The correct answer is (D)
Strong nuclear forces are responsible for holding together the nucleus of an atom; weak nuclear forces are involved when certain types of atoms break down.
7.3 kg of copper sits at a temperature of 38 degrees F. How much heat is required to raise its temperature to 865 degrees F? The specific heat of copper is 385 J/kg- degree C. Submit your anwser in exponential form.
The heat Q needed to increase the temperature of a sample with mass m and specific heat c by an amount ΔT is:
[tex]Q=mc\Delta T[/tex]On the other hand, a change in temperature in Farenheit is related to a change in temperature in Celsius as:
[tex]\Delta T_C=\frac{5ºC}{9ºF}\Delta T_F[/tex]Replace m=7.3kg, c=385J/(kgºC), as well as the final and initial temperatures to find the heat required to raise the temperature of the sample of Copper:
[tex]Q=(7.3\operatorname{kg})(385\frac{J}{\operatorname{kg}ºC})(865ºF-38ºF)[/tex]Since the specific heat is given in units of Joules per kilogram per degree Celsius, introduce the factor 5ºC/9ºF to write the change in temperature in degrees Celsius:
[tex]\begin{gathered} Q=(7.3\operatorname{kg})(385\frac{J}{\operatorname{kg}ºC})(865ºF-38ºF)\times\frac{5ºC}{9ºF} \\ =1,291,268.611\ldots J \\ \approx1.3\times10^6J \end{gathered}[/tex]Therefore, the amount of heat required to raise the temperature of the 7.3 kg of Copper sample from 38ºF to 865ºF, is 1.3*10^6 Joules.
Question 5 of 10Which phrase is the best definition of matter?A. A substance that cannot be divided into smaller piecesB. The smallest piece of a chemical compound that retains theproperties of the compoundC. Something that occupies a volume of space and also has massD. A substance that can change in both volume and shape
Something that occupies a volume of space and also has mass is the best definition of matter.
everything that occupies space and has mass. Matter, in the form of atoms, which are made up of protons, neutrons, and electrons, makes up all physical objects. Greek philosophers Democritus (470–380 BC) and Leucippus are credited with originating the notion that matter was composed of constituent parts or particles (490 BC).
Atoms form the basis of matter. One proton makes up the most fundamental atom, the protium isotope of hydrogen. Atoms are composed of protons, neutrons, and electrons, however these particles are based on fermions. Despite fitting certain definitions of matter, quarks and leptons aren't commonly regarded as types of matter. It's easiest to just declare that matter is made up of atoms at the majority of levels.
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A small electric motor produces a force of 3 N that moves a remote-control car 4 m every second. How much power does the motor produce?
f = force = 5N
v = velocity = 4m/s
p = power
p = force * velocity
Replacing:
p= 3N * 4 m/s = 12W
The motor produces 12W of power.
how do i convert 5.6 kg in hg?
5.6 kg can be converted to hg using unitary method .
The unitary method is a process of finding the value of a single unit, and based on this value
here
kg = kilogram
hg = hectogram
1 kg = 10 hg
using unitary method , we can write
5.6 kg = 5.6 * 10 hg = 56 hg
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I’m not sure how to even begin this problem. I know it’s an equilibrium problem
m= 27 kg
angle = 40°
mg = 27 x 9.8 = 264.6 N
Fty . x1 - mg (x1/2 ) = 0
Fty =
A man attempts to move a truck by pushing it, but he can't move it. Describe the work done by the man.
If a man tried to move a truck by pushing it, but he is not able to move it, then the work done by the man will be equal to zero.
What is Work?In physics, the word "work" involves measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied with in the direction of the displacement. The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.
As per the given question,
The man tries to move the truck, but he is not able to move. It means that the total displacement is zero, which means according to the formula of work done the work is also zero.
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On Earth, the gravitational force of a robotic helicopter is 1.8 kilograms. What is the helicopter's gravitational
force on Mars?
On Earth, g = 9.8m/s²
On Mars, g = 3.71 m/s².
(1 point)
O 6.68 N
O 0.49 N
O 17.64 N
O2.64 N
The gravitational force of the robotic helicopter of mass 1.8 kg on Mars is 6.68 N.
What is gravitational force?Gravitational force is the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface.
To calculate the helicopter's gravitational force, we use the formula below.
Formula:
F = mg'............ Equation 1Where:
F = Gravitational force of the helicopterm = Mass of the helicopterg' = Acceleration due to gravity of MarsFrom the question,
Given:
m = 1.8 kgg' = 3.71 m/s²Substitute these values into equation 1
F = 1.8×3.71F = 6.678F ≈ 6.68 NHence, the helicopter's gravitational force on Mars is 6.68 N.
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Question 1: A skateboarder is traveling at 2 m/s when she reaches a downwardincline and accelerates at 3 m/s/s for 4 seconds. What speed is she traveling atthe end of those 4 seconds?O 14 m/s6 m/sI do not know how to solve this question.24 m/sO 10 m/s
Given:
The initial velocity of the skateboarder is
[tex]v_i=\text{ 2 m/s}[/tex]The acceleration is
[tex]a=\text{ 3 m/s}^2[/tex]The time taken is t = 4 s
Required: The final speed of the skateboarder.
Explanation:
The final velocity can be calculated by the formula,
[tex]v_f=v_i+at[/tex]On substituting the values, the final velocity will be
[tex]\begin{gathered} v_f=2+(3\times4) \\ =14\text{ m/s} \end{gathered}[/tex]Final Answer: The final velocity of the skateboarder after 4 s is 14 m/s.
A racecar travels around a circular track of radius 400 meters at constant speed. If the car's acceleration is 4 m/s2, what is its speed?
We are given that a car travels in a circular motion with an acceleration of 4m/s^2. We are asked to determine the speed of the motion. To do that we will use the following formula:
[tex]a=\frac{v^2}{r}[/tex]Where:
[tex]\begin{gathered} a=\text{ acceleration} \\ v=\text{ velocity} \\ r=\text{ radius} \end{gathered}[/tex]We will solve for the velocity first by multiplying both sides by "r":
[tex]ar=v^2[/tex]Now we take the square root to both sides:
[tex]\sqrt[]{ar}=v[/tex]Now we plug in the values:
[tex]\sqrt[]{(4\frac{m}{s^2})(400m)}=v[/tex]Now we solve the operations:
[tex]40\frac{m}{s}=v[/tex]Therefore, the velocity of the car is 40 meters per second.