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andrey2020 [161]
3 years ago
9

Two asteroids with masses 5.34*10^3 kg and 2.06*10^4 are separated by distance of 5,000 m. What is the gravitational force betwe

en the astroids? Newtons law of gravitation is F gravity =Gm1 m2/ r^2. The gravitational constant G is 6.67*10^-11 N*M^2/kg^2
A. 1.4*10^-6N
B.4.00 N
C. 1.24*10^32N
D. 2.93*10^-10N

I NEED HELP ASAP
Physics
2 answers:
alex41 [277]3 years ago
8 0
The answer is D. 2.93*10^-10N
AveGali [126]3 years ago
4 0

Answer:

D. 2.93*10^-10N

Explanation:

A.pex

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Jen pushed a box for a distance of 80m with 20 N of force. How much work did she do?
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The answer is 1,600 J.

A work (W) can be expressed as a product of a force (F) and a distance (d):
W = F · d<span>

We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
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W = 20 kg*m/s² * 80 m
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We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat
Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)

Here, mw is the mass of water

Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal

Now, you need to calculate the heat released by the steam:

Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal

The work done by this engine is the difference between both heats:

W=Q_{1}-Q_{2}=62760-61960=800cal

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