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yuradex [85]
2 years ago
7

1. The mass of the Earth is 81 times the mass of the Moon. Using Newton's Law of Gravity, what would the effect of gravitation b

e if the mass of the Moon were increased by four times and the distance between Earth and
the Moon were cut in half?

Physics
1 answer:
wariber [46]2 years ago
8 0

Answer:

Correct

Explanation:

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calculate the work done to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it?
Charra [1.4K]
100N is how much work is needed 
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2 years ago
A rabbit is trying to cross the street. Its velocity v as a function of time t is given in the graph below where
Amiraneli [1.4K]

Answer:

2.5

Explanation:

8 0
2 years ago
A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla
ch4aika [34]

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

4 0
2 years ago
You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac
Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1  + \frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1  + \frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}

Stop sight distance= 81.88 m

5 0
2 years ago
What landforms form at convergent boundaries where two oceanic plates collide?
Bad White [126]
It forms mountains and sometimes islands.
4 0
2 years ago
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