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3 years ago

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A player kicks a football from ground level with an initial velocity of 27.0 30.0 above the horizontal Find the ball's maximum h

eight

1 answer:

stepladder [879]3 years ago

5 0

Answer:

Given

initial velocity (u) =27.030

Force of gravidity g) =9.8

Rtc maximum height Hmix =?

$sln \\ \ hmix = u {}^{2} \div 2g \\ 27.030 \div 2 \times 9 \\ \\ hmix = 38.025m$

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Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan

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Answer:

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

Explanation:

We are given that

Gravitational force= $F_g=\frac{Gm_Emr}{R^3_E}$

r=0,U(0)=0

We know that

Gravitational potential energy= $-\int F_gdr$

$U(r)=-\int\frac{Gm_Emr}{R^3_E}dr$

$U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C$

Substitute r=0 ,U(0)=0

$0=0+C$

$C=0$

Substitute the value

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The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

$F=mg$ (1)

where

m is the mass of the astronaut

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$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

$F=\frac{GMm}{r^2}$

When the astronaut is on the Earth's surface, $r=R$ (where R is the Earth's radius), so his weight is

$F=\frac{GMm}{R^2}=640 N$

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

$r'=3R+R=4R$

Therefore, the new weight is

$F'=\frac{GMm}{(4R)^2}=\frac{1}{16}\frac{GMm}{R^2}=\frac{F}{16}$

Which means that his weight has decreased by a factor 16: therefore, the new weight is

$F'=\frac{640}{16}=40 N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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