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nadya68 [22]
3 years ago
15

A chunk of ice with a mass of 1.2 kg melts and absorbs 3.33 x 10³J of energy. Which best describes what has happened in this sys

tem?
Physics
1 answer:
Maurinko [17]3 years ago
3 0

Answer:

- Its entropy increases.

Explanation:

Entropy is defined as a 'measure of the amount of energy in a physical system that cannot be used to do work.' It is also employed to denote randomness, disorder, or uncertainty of the arrangement/system. In the given system, the melting of ice denotes the 'increase in entropy' as the amount of energy unavailable to do work increases('absorbs 3.33 x 10³J of energy'). Thus, <u>this signifies that the entropy increases with a rise in temperature as it allows the substance to have greater kinetic energy</u>.

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Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown
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Answer:

Explanation:

Actual weight, Wo = 5 N

Apparent weight, W = 4.5 N

density of water = 1 g/cm^3 = 1000 kg/m^3

density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3

Buoyant force = Actual weight - Apparent weight

Volume x density of water x g = 5 - 4.5

V x 1000 x 9.8 = 0.5

V = 5.1 x 10^-6 m^3

Weight of gold = Volume of gold x density of gold x gravity

W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N

As W' is less than W so, it is not pure gold.

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Which items do NOT describe possible solutions to air pollution?
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Answer:

2, 6

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A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
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Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

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3 years ago
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