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jekas [21]
3 years ago
5

The football player running toward the goal line has

Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0

Kinetic energy.

Kinetic energy is the type of energy observed in moving objects. In this case the football player is running, ie moving, so he/she must have kinetic energy.

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A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0
ValentinkaMS [17]

M = mass of aluminium = 1.11 kg

c_{a} = specific heat of aluminium = 900

T_{ai} = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

c_{w} = specific heat of water = 4186

T_{wi} = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M c_{a} (T_{ai} - T) = m c_{w} (T - T_{wi} )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

7 0
3 years ago
During a medieval siege of a castle, the attacking army uses a trebuchet to heavy stones at the castle the trebuchet launches th
vlada-n [284]

Answer:

Explanation:

The question relates to time of flight of a projectile .

Time of flight = 2 u sinθ / g

u is speed of projectile , θ is angle of projectile

= 2 x 48.5 sin42 / 9.8

= 6.6 seconds  .

Maximum height attained

= u² sin²θ / g

= 48.5² sin²42 / 9.8

= 107.47 m .

7 0
2 years ago
You are a traffic accident investigator. You have arrived at the scene of an accident. Two cars of equal mass (1,000 kg each) we
kumpel [21]
6. Is 50 m/s that’s the answer
4 0
3 years ago
Is velocity a scalar or a vector
gizmo_the_mogwai [7]
Velocity<span> is a</span>vector<span> quantity; it is direction-aware.</span>
5 0
3 years ago
Read 2 more answers
In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

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
3 years ago
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