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lilavasa [31]
2 years ago
10

If a 50 KG object is at a location 25,600 km from Earth's Center, what is the gravitational force exerted by the objects on Eart

h? In what direction does that force act? Support your answer with evidence.
Physics
1 answer:
masya89 [10]2 years ago
5 0

Answer: The force exerted by the object on Earth is 30.4 N, and the direction is towards the object.

Explanation:

The gravitational force between two objects of mass M1 and M2, that are at a distance R between them is written as:

F = G*(M1*M2)/R^2

Where G is the gravitational constant, such that:

G = 6.674*1^(-11) m^3/(kg*s^2)

We know that M1, the mass of the object, is 50kg

R is the distance, in this case, is 25,600km

But this needs to be written in meters, remembering that:

1km = 1000m

Then:

25,600km = (25,600*1000)m = 25,600,000 m

And M2 is the mass of Earth, which is:

M2 = 5.972*10^(24) kg

Replacing all of those in the force equation we get

F = (6.674*1^(-11) m^3/(kg*s^2))*(50kg*5.972*10^(24) kg)/( 25,600,000 m)^2

F = 30.4 N

We know that the gravitational force is attractive, then the direction in which this force acts is towards the 50kg object.

Now, remember the second Newton's law is:

F = M*a

Then the acceleration that the object causes on Earth is:

30.4N = ( 5.972*10^(24) kg)*a

a = 30.4N/( ( 5.972*10^(24) kg)) = 5.09*10^(-24) m/s

This acceleration is almost despreciable.

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3 years ago
A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ s
Bingel [31]

Explanation:

According to the law of conservation of energy ,    

             Potential energy = kinetic energy

   mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}

                  I = \frac{mr^{2}}{2}

          \omega = \frac{v}{r}

     mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]

     mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]

             g \times h = \frac{3}{4} \times v^{2}

             9.8 \times 4.2 = \frac{3}{4} \times v^{2}

                  v = 7.4 m/s

thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.

5 0
3 years ago
Kirchhoff's Rules When applying Kirchhoff's rules, one of the essential steps is to mark each resistor with plus and minus signs
luda_lava [24]

Answer:

d. R4

Explanation:

Generally, the flow of current is always from the positive sign to the negative sign. In the resistors R1, R2, and R3, the direction of flow of current is from the positive sign to the negative sign. However, in the resistor R4, the direction of the flow of current is different from the conventional method. Therefore, the resistor R4 is marked wrongly.

8 0
2 years ago
Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co
aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s

The angular speed of the motion is 2.21433 rad/s

4 0
3 years ago
3. What is the relationship between the kinetic energy of molecules in an object and the object’s temperature? a. As the tempera
evablogger [386]

I think the answer A since temperature is the average kinetic energy of the molecules, so increasing temperature must increase kinetic energy

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