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Advocard [28]
3 years ago
9

A person standing on the surface of the earth has a weight of 650 N. What is its weight at a distance of two earth radii from th

e surface? If the person is at a distance of two earth radii from the surface, he/she is at a distance of three earth radii from the earth’s center. Since we have an inverse square law, three times the distance means 1/9th of the weight.
Physics
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

72.22 N

Explanation:

F = weight

m = mass of body

M = mass of earth

R = radius of earth

G = universal constant of gravitation

F_1= 650 N

F_1 = GMm/R^2

two earth radius above the surface of the earth:

F_2=  GMm/(3R)^2=  GMm/9R^2= F_1/9= 650/9

=72.22 N

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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
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The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

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3 years ago
Your teacher gives you a piece of cardboard with two pinholes on it, saying that they are separated by 205 μm ± 3%. In order to
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Answer:

m,lkj,mkn,njn

Explanation:because she is telling you to do a project

7 0
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From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ
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Answer:

55.8 m/s

Explanation:

f = Actual frequency of sound emitted by car

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the electric conductivity of gold is very high

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A forklift pushes a box with a force of 500 N across the floor for a distance of 5.0 m, then turns around and pushes with the sa
yanalaym [24]

Answer:

5,000J

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Work = Force x Distance

Distance back and forth is canceled out, so either the answer is + or -

5.0m + 5.0m = 10.0m

500N x 10.0m = 5,000J

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3 years ago
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