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Contact [7]
3 years ago
10

I REALLY NEED HELP PLEASE!!! CAN SOMEONE ANSWER THIS? At 0 ºC, some amount of energy is required to change 1 kg of water from a

solid into a liquid. If you had a 2 kg piece of ice, what effect would this have on the amount of thermal energy required to change the water from a solid to a liquid? A)It would require energy to be removed from the 2 kg piece of ice. The larger piece of ice already has more total energy than the smaller piece of ice, so energy must be removed for the ice to become liquid. B)It would require less energy to change solid water into liquid water because the energy would spread through the ice more quickly and the ice already has a larger total amount of energy because it is larger than a 1 kg piece of ice. C)It would require more energy to change solid water into liquid water because there are more molecules in this larger piece of ice. D)It would still require the same amount of energy to change solid water into liquid water because the entire piece of ice would still gain the same amount of energy in each case.
Physics
2 answers:
Nezavi [6.7K]3 years ago
8 0

Answer:

C)It would require more energy to change solid water into liquid water because there are more molecules in this larger piece of ice.

Charra [1.4K]3 years ago
4 0

Explanation:

c. it would require more energy to change ........

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Matter that emits no light at any wavelength is called DARK MATTER.

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3 years ago
In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t
Vitek1552 [10]

Answer:

So the ratio will be \frac{T_L}{T_H}=-0.171

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So Q=450j

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency \eta =\frac{W}{Q}=\frac{290}{450}=0.6444

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine =\frac{0.6444}{0.55}=1.171

Efficiency of Carnot engine is \eta =1-\frac{T_L}{T_H}

1.171 =1-\frac{T_L}{T_H}

\frac{T_L}{T_H}=-0.171

3 0
3 years ago
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Earth is about 93 million miles from the Sun but only 239,000 miles from the Moon. Likewise, the Sun has a mass that is over 20
Yakvenalex [24]

Answer:

Gravity is dependent on the mass of two bodies and the distance between them. There is a strong gravitational attraction between Earth and the Moon because they’re relatively close to one another. There is a strong gravitational attraction between Earth and the Sun because the Sun is so massive

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3 years ago
Please help on this one?
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D. An image that is smaller than the object and is behind the mirror

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4 years ago
A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of
Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.  

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

K_1 = K_2 + P + W

then, the work done by the engine is:

W = K_1 - K_2 - P

The formulas for the Kinetic and potential energy are:  

K=\frac{1}{2}mV^2\\P=mgh

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh

Replacing the values:

W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

t=\frac{3200m}{20m/s}=160s

the power delivered by the car's engine was:

power=\frac{work}{time}=\frac{1690400J}{160s}=10565W

8 0
4 years ago
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