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Vikentia [17]
3 years ago
9

3. An automobile accelerates 1.77 m/sover 6.00 s to reach the freeway speed at

Physics
1 answer:
Effectus [21]3 years ago
4 0

The initial speed of the automobile is 49.84km/hr

<u>Explanation:</u>

Given:

Acceleration, a = 1.77 m/s²

Time, t = 6s

Final speed, v = 88 km/h

                     v = 88 X 0.278 m/s

                     v = 24.464 m/s

Initial speed, u = ?

We know,

v = u + at

On substituting the value in the formula we get:

24.464 = u + (1.77 X 6)

24.464 = u + 10.62

u = 24.464 - 10.62 m/s

u = 13.844 m/s

Converting u = 13.844 m/s to km/hr

1 m/s = 3.6 km/hr

13.844 m/s = 13.844 X 3.6 km/hr

u = 49.84 km/hr

Therefore, the initial speed of the automobile is 49.84km/hr

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At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
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Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

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A student stretches an elastic band by 0.8 m in 0.5 seconds. The spring constant of the elastic band is 40 N/m. What was the pow
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Answer:

The power exerted by the student is 51.2 W

Explanation:

Given;

extension of the elastic band, x = 0.8 m

time taken to stretch this distance, t = 0.5 seconds

the spring constant, k = 40 N/m

Apply Hook's law;

F = kx

where;

F is the force applied to the elastic band

k is the spring constant

x is the extension of the elastic band

F = 40 x 0.8

F = 32 N

The power exerted by the student is calculated as;

P = Fv

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F is the applied force

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