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horsena [70]
3 years ago
5

The force of gravity between two objects depends on ___________.

Physics
1 answer:
murzikaleks [220]3 years ago
4 0
The answer is D. the mass of the two objects and the distance between them
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Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0
3 years ago
What are four minerals that are commonly taken from ores
Umnica [9.8K]
If it helps or doesn't I'm sorry, but if you even played the game Minecraft just remember it. 

Gold, silver, coal, and iron come from ores. 
5 0
3 years ago
A 4kg brick is dropped from the top of a building whose hight is 30m.what is the velocity with which it reaches the ground​
miss Akunina [59]

Answer:

=24.25 ^−1

Explanation:

Let   and   be initial and final velocity of the body respectively,  

be acceleration due to gravity ( 9.8^−2 ),  ℎ be the height of the body.

=0 ^ −1

ℎ=30

we know that, ^2−^ 2=2ℎ

^2=2∗9.8∗30

^2=588

=24.25 ^−1

4 0
3 years ago
What is the main reason for using a data table for collecting data?
Luba_88 [7]

Answer:

to make calculation more easy to get

Explanation:

if you are using chart or calculate Thermodynamic problems you will not never solve this problem with out using data table for thermodynamic

7 0
3 years ago
A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff
Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0
2 years ago
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