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IrinaK [193]
3 years ago
12

Which of the following statements is true? The melting and freezing points of a substance are the same. The melting and boiling

points of a substance are the same. The freezing and boiling points of a substance are the same. The heat of fusion and the heat of vaporization for a substance are the same.
Physics
2 answers:
insens350 [35]3 years ago
4 0
<span>The melting and freezing points of a substance are the same.

Substances melt and freeze at the same temperature. Hope this helps! :)</span>
ZanzabumX [31]3 years ago
4 0

Answer:

The melting and freezing points of a substance are the same

Explanation:

Melting occurs when the state of a substance changes from solid to liquid, while freezing occurs when the state of a substance changes from liquid to solid. Therefore, melting and freezing are one the inverse process of the other: it means that they occur at the same temperature.

For instance, the melting point of ice (at standard pressure) is 0^{\circ}, and the freezing point of water is at the same temperature, 0^{\circ}.

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butalik [34]
Makes no sense get a better question
4 0
3 years ago
If we add 50 Joules of thermal energy to a heat engine, and that heat engine does 30 Joules of work, how much thermal energy is
Natalka [10]

Answer:

The correct answer should be

A. 20 Joules

Explanation:

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7 0
3 years ago
A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0
3 years ago
I have no idea what to do. Plz help!
aleksklad [387]
Work is (force) times (distance). For Amy, you know both of them, and you can easily multiply them to find the amount of work. For Joe, the distance is zero, which should tell you all you need to know.
7 0
3 years ago
PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton b
GuDViN [60]

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

3 0
3 years ago
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