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3 years ago

A bus slows down uniformly from 16.62 m/s to 0 m/s in 10.39 s. How far does it travel before stopping?

Physics

1 answer:

antiseptic1488 [7]3 years ago

6 0

Answer:

bro ily

Explanation:

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A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on

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Answer:

(a) 81.54 N

(b) 570.75 J

(d) 0 J, 0 J

(e) 0 J

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J

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3 years ago

An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

zimovet [89]

Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

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A bus slows down uniformly from 16.62 m/s to 0 m/s in 10.39 s. How far does it travel before stopping?​

A bus slows down uniformly from 16.62 m/s to 0 m/s in 10.39 s. How far does it travel before stopping?