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Colt1911 [192]
3 years ago
5

Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of

your current weight. Show all work leading to your answer OR describe your solution using 3 -4 complete sentences.
Physics
1 answer:
FrozenT [24]3 years ago
6 0
Gravity obeys the inverse square law.  At 6400 km above the center of the Earth (Earth's surface) you weigh x.  Twice that reduces your weight to 1/4th.  Four times that height reduces your weight to 1/16th.  4 times 6400 km is 25,600 km.  But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer:  19,200 km.

Incidentally, it doesn't exactly work the opposite way.  At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
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Your cousin has looked after you your whole life and you always looked up to him. Today he told you he sells drugs; " That's whe
tatuchka [14]

Answer:

Honestly, you should have a firm resolve not to go into drug peddling.

Politely tell him you are not interested in the business, that you are most concerned about your future and the career path you have chosen for yourself already.

He will definitely feel bad, but you can go on to advise him.

You can also advise him to stop and highlight the dangers of doing drugs.

Explanation:

There are a whole lot of disadvantages involved in the drug business,

in fact, the disadvantages out weights the advantages in the long run.

say if one is caught and sent to a life long sentence, or if one is caught up with the effect of the use of hard drugs and the health implication, to the extent that one looses his/her sanity, you see it won't make any sense anymore doing drugs in the long run. so friend you can advise you cousin to stop and thinker of good businesses that cold fetch him cool money, without the law chasing him, with that he can be a contributor to economic growth by paying his taxes too

3 0
3 years ago
A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?
iren2701 [21]

Answer:

K.E=29.403125J

Explanation:

From the question we are told that

Mass M=100

Height 50-20=30m

Generally the equation for velocity before impact is is is mathematically given by

v=\sqrt{2gh}

v=\sqrt{2*9.8*30}

v=24.25

Generally the equation for Kinetic Energy is is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*100*(24.25)^2\\

K.E=29403.125J

K.E=29.403125J

8 0
3 years ago
You walk to the north, then turn 90° to your left and walk another How far are you from where you originally started?
rodikova [14]

Whatever distance north and then west you walked, you are then

(1.41 x that distance)

northwest of where you started.

3 0
3 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
A 60-kg block initially at rest on a frictionless horizontal surface is acted upon by a force of 3.0 N for a distance of 7.0 m.
Schach [20]

Answer:

Explanation:

Kinetic energy generated = work done by force = force x displacement

= 3 x 7 = 21 J

5 0
3 years ago
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