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Aleonysh [2.5K]
3 years ago
5

Moon effect. some people believe that the moon controls their activities. if the moon moves from being directly on the opposite

side of earth from you to being directly overhead, by what percentage does (a) the moon's gravitational pull on you increase and (b) your weight (as measured on a scale) decrease? assume that the earth–moon (center-to-center) distance is 3.82 × 108 m, earth's radius is 6.37 × 106 m, moon's mass is 7.36 × 1022 kg, and earth's mass is 5.98 × 1024 kg.
Physics
1 answer:
Anon25 [30]3 years ago
5 0
Actually thw moon doesnt pull on you...the earth is 4 times higher with a gravitational pull of 6 times more than the moon.
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A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) H
Wewaii [24]

Answer:

(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s

Explanation:

Given data

v_{o}=7.14m/s\\\alpha =36.5^o

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2

Using equation of motion

v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s

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3 years ago
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The correct answer is

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