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lana [24]
3 years ago
9

A Car accelerates from 4 m/s to 16 m/s in 4 seconds. What is the car's

Physics
2 answers:
klemol [59]3 years ago
5 0

Answer:

4 m/s^2

that is the right acceleretion a car

irakobra [83]3 years ago
5 0

Answer:

I believe that the answer is 3 m/s^2

Explanation:

since acceleration is final velocity minus initial velocity divided by time, we can plug in known values and solve.

a=(f-i)/t

a=(16-4)/4

a=12/4

a=3

3m/s^2 is the acceleration

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andreev551 [17]
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.

When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.

a)

Magentic force, F = q*v*B

q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T

Centripetal force, F = m*Ac = m * v^2 / R

where,

Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit

Now equal the two forces: q*v*B = m * v^2 / R => R =  m*v / (q*B)

=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]

=> R = 0.114 m

b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.

R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]

=> R = 10.4 m

 

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4 years ago
Please help with the working of this question
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Answer:

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