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2 years ago

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The mass of earth 6*10^24kg and radius is 400kg Now find the value of acceleration due to gravity when a object is 3600km from t

he earth surface

1 answer:

vlada-n [284]2 years ago

5 0

I don’t worry wewwwww it is a good time to get it done lol lol i don’t worry about it lol lol i lol

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an object is acted on by a drag force with a magnitude that is proportional to the speed. the object accelerates downward at 3.0

Nutka1998 [239]

The terminal speed of the object falling down is 66.67 m/s.

The terminal speed acquired by the body when,

Weight of the body = Drag force of the body

It is given,

Drag force is directly proportional to the speed,

So,

F = CV

Where F is drag force,

V is the speed,

C is the constant,

So, it can be written as C = F/V.

The weight of the body = mg

The weight of the body = 10m

M is the mass and g is the acceleration due to gravity,

The drag force when the speed is 20m/s.

Drag force = ma

a is the acceleration during the drag force which is given to be 3m/s²,

Drag force = 3m

Now we can write,

F₁/V₁ = F₂/V₂

F₁ is the drag force at 20m/s speed.

F₂ is the weight of the body and V₂ is the terminal speed,

Now, it can be written,

3m/20 = 10m/V₂

V₂ = 66.67 m/s.

So, the terminal speed is 66.67m/s.

To know more about terminal speed, visit,

brainly.com/question/14605362

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7 0

1 year ago

A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while los

mafiozo [28]

Answer:

Option B

Change in entropy of the process is $\Delta S= Rln(\frac{P_{1}}{P_{2}})$

Explanation:

The entropy of a system is a measure of the degree of disorderliness of the system.

The entropy of a system moving from process 1 to 2 is given as

$\Delta S = \int\limits^2_1 {\frac{\delta q}{T}}$

recall from first law, $\delta q =du +Pdv$

hence we have, $\Delta S = \int\limits^2_1 {\frac{du +Pdv}{T}}$

since the process is isothermal, $du= 0$

this gives us $\Delta S = \int\limits^2_1 {\frac{Pdv}{T}}$

integrating within the limits of 1 and 2, will give us

$\Delta S = R ln (\frac{V_{2}}{V_{1}})$

also from ideal gas laws,

$\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}$

hence we have $\Delta S = R ln (\frac{P_{1}}{P_{2}})$

This makes the correct option B

4 0

3 years ago

Read 2 more answers

How should the magnetic field lines be drawn for the magnets shown below?

sergeinik [125]

Option B is the correct answer that show how magnetic field lines should be drawn for the magnets shown in the figure.

<h3>What is Magnetic Line of Force ?</h3>

The Magnetic Line of Force of a magnet is defined as the line along which a free N - pole would tend to move if placed in the field of a line such that the tangent to it at any point gives the direction of the field at that point.

When the two unlike poles are placed to each other, there will be attraction. And when the two like poles are placed to each other, there will be repulsion. The reason is that the line of force tend to move from the north pole to the south pole.

From the given diagram, the two magnets are of the same south pole. They are of like pole and there will be repulsion between the two magnets.

Therefore, Option B is the correct answer that show how magnetic field lines should be drawn for the magnets shown in the figure.

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5 0

2 years ago

What Law and explain why.<br><br> An airplane turns on its jets to accelerate down a runway.

Kaylis [27]

Answer:Newton's 2nd law

Explanation:Newton's 2nd law also states that the rate at which an object changes speed is proportional to the force that is exerted. Engines provide thrust and accelerate a plane forward along the runway. If the engines supply a small force, only a small acceleration will result.

5 0

2 years ago

An object is moving in the plane according to these parametric equations:

aniked [119]

A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π

b. vy = 4π cos (4πt + π/2)
vy = 0

c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]

d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax

h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax

i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)

h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.

5 0

3 years ago

Read 2 more answers