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Zina [86]
3 years ago
11

A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i

s 120 kg. The players stick together and are at rest after the collision. Find the speed of the second player, assuming the speed of light is 3.0 m/s.
Physics
1 answer:
Ket [755]3 years ago
7 0

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

m_1v_1+m_2v_2=(m_1+m_2)V

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

v_2=-\dfrac{m_1v_1}{m_2}

v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}

v_2=-1.47\ m/s

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

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When exposed to a radioactive source which emits 1.2-MeV gamma-rays, a particular material is found to have a half-value thickne
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Answer:E

Explanation:

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Shielding effect can be measured by measuring the fraction of gamma rays blocked by shield. If certain thickness will able to block half the radiation then to block 75% radiation we need to add same amount of thickness in order to block the remaining radiation.

i.e. \frac{E}{2} fraction is blocked by 10 cm thickness

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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

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