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melomori [17]
2 years ago
9

How much heat is given off by 1,000 kg of lead when it solidifies at its melting point?(see the table)

Physics
1 answer:
yuradex [85]2 years ago
7 0
At the melting point the heat given off is the latent heat of fusion. This occurs with at the melting point. It is not associated with temperature change.

All melting and solidification or freezing occur at the melting point temperature.

Q = mL      (This formula is associated with fusion or vaporization process)

Where Q = Heat associated with the melting process
m = mass in kg,
L = Latent heat of fusion in J/kg

For lead so we select the latent heat of fusion for lead.  Mass = 1000 kg

Q = 1000 kg * 2.04 x 10^4 J/kg

Q = 2.04 * 10^7  J

Option D.
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jonny [76]

Explanation:

Given that,

Mass of the shopping cart, m_s=21\ kg

Initial speed of the shopping cart, u_s=4\ m/s

Mass of the box, m_b=7\ kg

Initial speed of the box, u_b=0 (at rest)

They stick together and continue moving at a new velocity. It is a case of inelastic collision.

(a) The momentum of the shopping cart before the collision is given by :

p_s=m_s\times u_s\\\\p_s=21\times 4\\\\p_s=84\ kg-m/s

(b) The momentum of the box before the collision is given by :

p_b=m_b\times u_b\\\\p_b=7\times 0\\\\p_b=0

(c) The velocity of the combined shopping cart/box wreckage after the collision is given by using the conservation of momentum as :

m_su_s+m_bu_b=(m_s+m_b)V\\\\V=\dfrac{m_su_s+m_bu_b}{(m_s+m_b)}\\\\V=\dfrac{21\times 4+0}{(21+7)}\\\\V=3\ m/s

Hence, this is the required solution.

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2 years ago
If you throw an apple towards the sky why its going up in spite of gravitational pull?
AlexFokin [52]

Answer:

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The Force You Put Behind Said Apple Is Greater Than That Of Gravitational Pull

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2 years ago
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3 years ago
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As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte
Klio2033 [76]

Answer:

 r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

Explanation:

For this exercise we must use the principle of conservation of energy

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final point. The point where the proton is stopped (v = 0)

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where the potential is

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Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

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          r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

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