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kap26 [50]
3 years ago
12

A child is pushing a merry-go-round. The angle through which the merry-go-round has turned varies with time according to θ(t)=γt

+βt3θ(t)=γt+βt3, where γ=γ= 0.450 rad/srad/s and β=β= 1.10×10−2 rad/s3rad/s3. Part A Calculate the angular velocity of the merry-go-round as a function of time. Express your answer in terms of the variables ββ, γγ, and tt. ωz(t)ωz(t) = nothing rad/srad/s SubmitRequest Answer Part B What is the initial value of the angular velocity? ωzωz = nothing rad/srad/s SubmitRequest Answer Part C Calculate the instantaneous value of the angular velocity ωzωz at t=t= 5.25 ss . ωzωz = nothing rad/srad/s SubmitRequest Answer Part D Calculate the average angular velocity ωav−zωav−z for the time interval t=0t=0 to t=t= 5.25 ss .
Physics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

hufiui

fihgpfghlfikgergkfkjhfkhjgkffhhh

Explanation:

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A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same
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6 0
4 years ago
Will give 45 points to who ever solves this
Lubov Fominskaja [6]

Answer:

1.  230 kg...

W = m * g

W= 230 kg  * 9.81 m/s^2

<u>W=  2256,3 N</u>

     m= 230kg ,   W = 2256,3 N , g= 9.81 m/s^2

2.  887 N

W= m * g

887 N = m * 9.81 m/s^2

<u>m= 90,42 kg</u>

     m= 90,42 kg,  W = 887 N ,  g= 9.81 m/s^2

3.  420 kg

W= m * a

w= 420 kg *  9.81 m/s^2

<u>w=</u><u>  </u><u>4120,2 N</u>

     m=  420 kg , W = 4120,2 N , g= 9.81 m/s^2

4. Determine the gravity on Pluto where a 15 kg object weighs 55.5N.

w = m * g

55.5 N = 15 kg * g

<u>a=  3,7 m/s^2</u>

     m = 15 kg , W= 55.5 N , g=  3,7 m/s^2

7 0
3 years ago
Read 2 more answers
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