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IrinaVladis [17]
3 years ago
13

In my solar system, we have a planet that is the innermost to our star that is exactly like the innermost planet in your solar s

ystem. The planet has a very large impact crater with a basin that covers a large region of the planet's surface, but few smaller craters have formed on top of it. What can one conclude from this?
Physics
1 answer:
julia-pushkina [17]3 years ago
3 0

Answer:

1. The planet doesn't have a thick enough atmosphere.

2. There have been multiple impacts on the planet.

Explanation:

As the planet is very close to the star, there is high possibility that it will not have an atmosphere. Just like Mercury doesn't have one. Presence of a very large crater with basin indicates that in the past a huge body had hit the planet and thus creating the crater with basin. Also, it must be very old.

Second observation that is given is the presence of smaller craters in the basin. This indicates impact craters created by smaller objects. If the planet had an atmosphere, these smaller objects would not be able to penetrate and reach the surface. Thus presence of these smaller crater indicate towards the planet not having any atmosphere.

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A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh
Jet001 [13]

Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature =  80.0ºC

to find out

heat used to rise temperature of  pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature    ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature    ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0
3 years ago
Which is greater the attraction of the earth for 1 kg of aluminum or aluminum or attraction of 1kg of aluminum for the earth?
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Answer:

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