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FinnZ [79.3K]
2 years ago
14

A race car goes from a complete stop at the start line to 150 miles per hour in 5 seconds. What is its acceleration? Show your w

ork.
Physics
1 answer:
qwelly [4]2 years ago
3 0

Answer:

Explanation:

150/5  = 30

30mph per 1 second

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Suppose the sun shrank in size but its mass remained the same. what would happen to the orbit of earth?
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3 years ago
A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr
schepotkina [342]

Answer:

a

The New angular speed is  w_f = 2.034 rad/s

b

The Kinetic energy before the masses are pulled in is  KE_i = 3.101 \ J

c

The Kinetic energy after the masses are pulled in is   KE_f = 8.192 \ J

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

             The radius r =1.08m

              The  mass is m = 3.09\  kg

              The  angular speed w = 0.770 \ rad/sec

  The moment of inertia of the system excluding the two mass I = 3.25 \ kg \cdot m^2

           New radius  r_{new} = 0.34m

             

Generally the conservation of angular momentum can be mathematical represented as

                         w_f = [\frac{I_i}{I_f} ]w_i .....(1)

Where w_f is the final angular speed

           w_i is the initial  angular speed

          I_i is the initial moment of inertia

           I_f is the final moment of inertia

Moment of inertia is mathematically represented as

                       I = m r^2

Where I is the moment of inertia

          m is the mass

           r is the radius

So the Initial moment of inertia is given as  

     I_i = moment \ of \  inertia \ of\  the  \ two \  mass \ + 3.25 \ kg \cdot m^2

     I_i = 2m r^2 + 3.25

The multiplication by is because we are considering two masses

    I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2

So the final  moment of inertia is given as  

     I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2      

Substituting these values into equation 1

         w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec                                                          

Generally Kinetic energy is mathematically represented in term of moment of inertia as

                       KE = \frac{1}{2} * I * w^2

Now considering the kinetic energy before the masses are pulled in,

                     KE_i = \frac{1}{2} * I_i * w^2_i

The Moment of inertia would be  I_i = 10.46 \ Kg \cdot m^2

  The Angular speed would be  w_i = 0.77 \ rad/s

Now substituting these value into the equation above

              KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J

Now considering the kinetic energy after the masses are pulled in,                      

               KE_f = \frac{1}{2} * I_f * w^2_f

The Moment of inertia would be  I_f = 3.96 \ Kg \cdot m^2

The Angular speed would be  w_f = 2.034 \ rad/s

Now substituting these value into the equation above

                        KE_f= \frac{1}{2} *(3.96)(2.034)^2  

                        = 8.192J        

8 0
3 years ago
Hurry please I don’t have long for this for a test !!!
Anna35 [415]
I think the Answer is b
7 0
3 years ago
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