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Sonbull [250]
4 years ago
13

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

rcuit breaker trips at t = 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between t = 0 and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?
Physics
1 answer:
kozerog [31]4 years ago
3 0

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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3 years ago
The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0
AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t  represented by function :

p(t)=t^{2} +2t-4

Now,

When  function p(t) = 11 , t will be

p(t)=t^{2} +2t-4

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0  

t = 3 , -5  

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = \frac{position}{time}

therefore to get the value of velocity from  function p(t)

we have to differentiate the function with respect to time

\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)

v(t) = 2t + 2  

where v(t) = velocity at that time

as t = 3 for  p(t)=11  

so ,

v(t) = 2t + 2  

v(t) = 2*3 + 2  

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

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3 years ago
Five men push a stalled car with an average force of 400 N per person. Find the mass of the
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The mass of the car is 2000 kg

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

In this problem, we have:

a=1 m/s^2 is the acceleration of the car

Each person applies a force of 400 N, and there are five men, so the total force applied is

\sum F = 5\cdot 400 N = 2000 N

Therefore, the mass of the car is:

m=\frac{\sum F}{a}=\frac{2000 N}{1 m/s^2}=2000 kg

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The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
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Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

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Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

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multiply each side by M_{o2}, so we have

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T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

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