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Deffense [45]
3 years ago
10

Earth's orbit around the sun is not perfectly circular. The nearest point of its orbit is 1.471E8 kilometers. The mass of the su

n is 1.989E30 kilograms. The mass of the earth is 5.97E24 kilograms. The universal gravitational constant is 6.6738E-11 N(m/kg)2. (a) What is the force exerted by the sun on the earth they are closest together? (b) What is the corresponding force of the earth on the sun?
Physics
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

a) -3.66233E22 N

b) 3.66233E22 N

Explanation:

Closest distance between Earth and Sun = 1.471E8 kilometers = r

Mass of Earth = 5.97E24 kilograms = m

Mass of the sun = 1.989E30 kilograms = M

a) Universal gravitational constant = 6.6738E-11 N(m/kg)2

F_{Mm}=-G\frac{Mm}{r^2}\\\Rightarrow F=-6.6738E-11 N \frac{5.97E24\times 1.989E30}{(1.471E11)^2}\\\Rightarrow F=-3.66233E22\ N

Force exerted by the sun on the earth they are closest together is -3.66233E22 N

b) The magnitude of the force that the earth will exert on the sun will be same but the direction will be different represented by the change of the sign

F_{mM}=-F_{Mm}\\\Rightarrow F_{mM}=-(-3.66233E22\ N)\\\Rightarrow F_{mM}=3.66233E22\ N

Force exerted by the earth on the sun when they are closest together is 3.66233E22 N.

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Given

The projectile is in air for a time of t=8 sec

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The time taken to reach the highest point is 4 sec

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The time taken is 4 sec.

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3 years ago
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3 years ago
A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s
Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is:   w_{c} =m_{c} g   and maximum tension is T=2.50 m_{c} g=1.41*10^4N

total mass and weight is :

M =m_{c}+ m_{b} =740kg+550kg=1290 kg

w_{M} =1.2650*10^4N

∑F_{y} =ma_{y}

T-M_{g} =Ma_{y}

a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg

=0.645m/s^2

B)

maximum acceleration

a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0

using y-y_{0} =v_{oy} t+1/2(a_{y} )t^2

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t=\sqrt{2(y-y_{0} )/a_{y} }

t=\sqrt{2(119m)/0.645m/s^2} =19.20s

6 0
3 years ago
E)
guapka [62]

Answer:

Pressure, P = 32666.66 Pa

Explanation:

It is given that,

Surface area of foot of Bimaba is 150 cm² or 0.015 m².

Her weight is 50 kg

We need to find the pressure does she exert on the ground, as she stands on  her one foot. The force acting per unit area is called pressure. It can be given by :

P=\dfrac{F}{A}\\\\P=\dfrac{mg}{A}\\\\P=\dfrac{50\times 9.8}{0.015}\\\\P=32666.66\ Pa

So, the pressure is 32666.66 Pa.

7 0
3 years ago
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