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ohaa [14]
2 years ago
13

Which of these actions would increase heat transfer between two objects?

Physics
2 answers:
sergejj [24]2 years ago
7 0

Answer:

b- increases the area of their contact

Explanation:

i did it in edge! ; )

Kaylis [27]2 years ago
3 0

Answer:

Increasing the area of their contact.

Explanation:

Just took the test and it was right.

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You push a coin across a table. The coin stops. How does this motion relate to balanced and unbalanced forces?
Snezhnost [94]
If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.
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Which describes why nonrenewable energy resources are used more frequently than renewable ones?
vodomira [7]
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3 years ago
The speed of light in a vacuum is approximately 2.99 × 108 m/s, and the speed of light through a piece of glass is approximately
exis [7]

Option (b) is correct.The index of refraction for the glass is 1.52

Explanation:

velocity of light in vacuum= C= 2.99 x 10⁸m/s

Velocity of light in glass = V= 1.97 x 10⁸m/s

The refractive index is given by n=\frac{c}{v}

n= 2.99 x 10⁸/1.97 x 10⁸m/s

n= 1.52

Thus the refractive index of glass is 1.52

6 0
3 years ago
Read 2 more answers
A piece of bismuth with a mass of 4.06 g 4.06 g gains 423 J 423 J of heat. If the specific heat of bismuth is 0.123 J / ( g ° C
Sholpan [36]

Answer: 846°C

Explanation:

The quantity of Heat Energy (Q) required to heat bismuth depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = 423 joules

Mass of bismuth = 4.06g

C = 0.123 J/(g°C)

Φ = ?

Then, Q = MCΦ

423 J = 4.06g x 0.123 J/(g°C) x Φ

423 J = 0.5J/°C x Φ

Φ = (423J/ 0.5g°C)

Φ = 846°C

Thus, the change in temperature of the sample is 846°C

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3 years ago
One runner in a relay race has a 1.50 s lead and is running at a constant speed of 3.25 m/s. The runner has 30.0 m to run before
yarga [219]
The second runner must run 3.3m/s. If the leading runner is 1.5 seconds ahead and there are 30m left, the second runner would need to run slightly faster than the lead in order to finish at the same time. To calculate this I did 30/1.5 which gave me 0.05. I added this onto the speed of the lead runner to get 3.3m/s :)
6 0
3 years ago
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