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sergey [27]
3 years ago
10

A boy and a girl are riding a merry- go-round which is turning ata constant rate. The boy is near the outer edge, while a girl i

s closer to the center. Who has the greater tangential acceleration? 1. the boy 2. the girl 3. both have zero tangential 25% try penalty acceleration Hints: 0,0 4. both have the same non-zero tangential acceleration
Physics
1 answer:
max2010maxim [7]3 years ago
3 0

Answer:

The correct answer is

1. the boy 2.

Explanation:

Since Tangential acceleration = v²/r  then the

v = ωr and v² =  ω²r² then the tangential acceleration = ω²r²/r = ω²r

This shows that  since ω is the same for both of them, the boy with greater r has the most acceleration.

The tangential acceleration a_t = α·r

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In a single wire, how much current would be required to generate 1 Tesla magnetic field at a 2 meter distance away from the wire
aev [14]

Answer:

12.56 A.

Explanation:

The magnetic field of a conductor carrying current is give as

H = I/2πr ............................... Equation 1

Where H = Magnetic Field, I = current, r = distance, and π = pie

Making I the subject of the equation,

I = 2πrH............... Equation 2

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3 years ago
In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
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Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

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