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bagirrra123 [75]
3 years ago
5

What has more momentum a large truck stopped at a stop sign or a motorcycle moving down the road.

Physics
1 answer:
Alona [7]3 years ago
4 0
<h2>I'm pretty sure it's</h2><h2>"<u>a motorcycle moving down the road.</u>"</h2><h3></h3><h3>The truck isn't moving so the truck cannot have momentum.</h3><h3>Momentum is basically "mass in motion."</h3><h3></h3><h3><em>Please let me know if I am wrong.</em></h3>
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500

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5. What is the density of 4.5 mL of a liquid that has a mass of 1.3 grams?
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A. 0.289g/mL

Explanation:

Using the equation for density which is d = m/v  or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.

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A battery has a terminal voltage of 12.0 V when no current flows. Its internal resistance is 2.0 Ω. If a 4.6 Ω resistor is conne
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Check attachment for solution

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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

5 0
2 years ago
Yashoda prepares some lime juice on a hot day. She adds 80 g of ice at a temperature of 0°C to 0.32 kg of lime juice. The temper
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Answer:

Explanation:

a )

hear energy required to melt 1 g of ice = 340 J ,

hear energy required to melt 80 g of ice = 340 x 80  J = 27220 J .

b ) energy gained by the melted ice ( water at O°C ) = m ct

where m is mass of water , s is specific heat and t is rise in temperature

= 80 x 4.2 x ( 8°C - 0°C)

= 2688 J .

c )

energy lost by lime juice = energy gained by ice and water

= 27220 J + 2688 J .

= 29908 J .

d )

Let specific heat required be S

Heat lost by lime juice = M S T

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= 6720 S

For equilibrium

Heat lost = heat gained

6720 S = 29908 J

S = 4.45 J /g °C .

4 0
3 years ago
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