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2 years ago

1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of

Physics

1 answer:

nevsk [136]2 years ago

3 0

Answer:

$4.8 \cdot 10^4 N$

Explanation:

The force of gravity acting on the satellite is given by:

$F=\frac{GMm}{r^2}$

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

$r=2.4\cdot 10^6 m$

Substituting into the equation, we find:

$F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N$

Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).

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The pure form of an element that is typically lustrous, malleable, and conducts heat and electricity is a/an ___________.

aliya0001 [1]

The pure form of an element that is typically lustrous, malleable, and conducts heat and electricity is a/an metal

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2 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the

PIT_PIT [208]

The drag force is directly proportional to the square of the velocity of motion of the object. So as the speed is doubled, the magnitude of drag force will get quadrupled.

Explanation:

Drag force is the opposing or resisting force acting on any object by the medium in which it is moving. So in this case, you are riding a bicycle, thus the medium can be considered as air.

The formula for calculating drag force is as below:

$\text {Drag force }=\frac{1}{2} \rho v^{2} C_{D} A$

Here, ρ is the density of the air molecules, v is the velocity or speed of the bicycle, CD is the drag coefficient and A is the area of the bicycle.

In the above equation, only the term velocity will be a varying quantity with respect to time and other quantities will remain constant throughout the single situation of riding of bicycle.

So, the equation can be,

$\text { Drag force }=k v^{2}$

Where ,

$k=\frac{1}{2} \rho C_{D} A$ (constant for this whole condition)

Now given the speed of bicycle increased from v to 2v, so the initial drag force will be

$N_{i}=k v^{2}$

After increase in speed, the final drag force will be

$N_{f}=k(2 v)^{2}$

$N_{f}=4 k v^{2}=4 N_{i}$

Thus, if the speed of the bicycle is doubled, the drag force will get increased by four times.

4 0

2 years ago

Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass

kirza4 [7]

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

5 0

2 years ago

What is the energy released in this B- nuclear reaction 2K-> 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca

Katyanochek1 [597]

Answer: The energy released in the given nuclear reaction is 3.526 MeV.

Explanation:

For the given nuclear reaction: