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sweet [91]
3 years ago
13

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

h's surface and a kinetic energy of 4200 MJ . Later in its orbit the satellite's potential energy is 5700 MJ . Part A Use conservation of energy to find its kinetic energy at that point.
Physics
1 answer:
serious [3.7K]3 years ago
5 0

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

E_{initial} = E_{final}

PE_{initial}+KE_{initial} = PE_{final}+KE_{final}

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}

KE_{final} = (5100+4200)-5700

KE_{final} = 3600MJ

Therefore the final kinetic energy is 3600MJ

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Aunt Matilda goes to a well and throws a penny straight down the well at 3.0 m/s. She hears a splash after 0.5 seconds. How deep
nevsk [136]

Answer : The correct option is (d) 2.73 m

Explanation :

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2

where,

s = distance or height = ?

u = initial velocity  = 3.0 m/s

t = time = 0.5 s

a = acceleration due to gravity = 9.8m/s^2

Now put all the given values in the above equation, we get:

s=(3.0m/s)\times (0.5s)+\frac{1}{2}\times (9.8m/s^2)\times (0.5s)^2

s=2.73m

Therefore, the correct option is (d) 2.73 m

8 0
3 years ago
When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5
torisob [31]

Answer:

Explanation:

We put the charges in the ascending order as follows

1.53 P

3.26 P

4.66 P

5.09 P

6.39 P

where P is equal to 10⁻¹⁹

we round off given charges as follows

1.53 P → 1.6 P

3.26 P → 3.2 P

4.66 P → 4.8 P

5.09 P → 4.8 P

6.39 P → 6.4 P

We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .

Hence this charge may exist on single electron.

8 0
3 years ago
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum ha
sineoko [7]

Answer:

Therefore, the moment of inertia is:

I=0.37 \: kgm^{2}

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

T=2\pi \sqrt{\frac{I}{Mgd}} (1)

Where:

  • I is the moment of inertia
  • M is the mass of the pendulum
  • d is the distance from the center of mass to the pivot
  • g is the gravity

Let's solve the equation (1) for I

T=2\pi \sqrt{\frac{I}{Mgd}}

I=Mgd(\frac{T}{2\pi})^{2}

Before find I, we need to remember that

T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s

Now, the moment of inertia will be:

I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}  

Therefore, the moment of inertia is:

I=0.37 \: kgm^{2}

I hope it helps you!

7 0
3 years ago
A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance
seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

4 0
3 years ago
A baseball has a mass of 145 g. A bat exerts a force of 18,400 N on the ball. What is the acceleration of the ball?
amid [387]

The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.

From the details given in the question, we are told that:

F = 18, 400N

M = 145 g = 145 / 1000 = 0.145 kg

A = ?

From the equation F = MA

A = F / M

A = 18,400 / 0.145 = 126,896.55 = 1.27 *10^5.

Therefore, the correct option is C.

3 0
3 years ago
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