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olya-2409 [2.1K]
3 years ago
8

A satellite in orbit is not truly traveling through a vacuum.It is moving through very, very thin air. Does the resulting airfri

ction cause the satellite to slow down?
b) Explain the similarities anddifferences between Newton’s law of universal gravitation andCoulomb’slaw.
Physics
1 answer:
defon3 years ago
4 0

PART A) Yes, the fact that there is a frictional force acting on the satellite generates a loss of energy due to friction. What causes satellite to diminish its orbit during its tour. In fact, many satellites have rectifier systems that allow them to position themselves and remain in their orbit for a long time to avoid being trapped by the Earth's gravity Force and fall into the atmosphere where they would probably be torn apart.

PART B) As a similarity, one could start by mentioning the structure of the two equations are similar and have their own constants who were responsible for supporting them. While the law of gravity speaks of the masses of the bodies the electrostatic law speaks of the charges of the bodies. For both the force is inversely proportional to the square of the distance that separates them.

However, the most notable difference between them is basically their statement. While one of the equations speaks about greavedad the other reflects the electromagnetic phenomena. It should be noted that the force of gravity is much weaker than the electromagnetic force and that the latter has the capacity of attraction and repulsion. While the gravitational force only that of attraction.

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What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and
gayaneshka [121]

Force applied on the car due to engine is given as

F_1 = 300 N towards right

Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

now the net force on the car will be given as

\vec F_{net} = \vec F_1 + \vec F_2

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

F_{net} = F_1 - F_2

F_{net} = 300 - 150

F_[net} = 150 N

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0
3 years ago
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7 0
3 years ago
a moving cart hits a spring, traveling at 5.0 m/s at the time of contact. at this instant the cart is motionless by how much is
viva [34]
This state of motionlessness occurs because all of the kinetic energy in the car is absorbed by the spring in the form of elastic potential energy. The mathematical representation is:
1/2 mv² = 1/2 kx²
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4 0
3 years ago
draw a velocity graph with Vi = 4 m/s and decreasing uniformly so that velocity at 2 seconds is 2 m/s and remaining constant fro
Shkiper50 [21]

For the velocity graph: start at 0s and 4m/s and draw a straight line to 2s and 2 m/s. Then draw a straight horizontal line to 4s and 2m/s

For the acceleration graph: start with a horizontal line from 0s and 2m/s/s to 2s and 2m/s/s. The draw another line from 2 s and 0m/s/s to 4 s and 0m/s/s

6 0
3 years ago
An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10
STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi(\frac{6894.76\ Pa}{1\ psi}) = 379212 Pa

\rho = density of water = 1000 kg/m³

Therefore,

v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

\frac{V}{t} = Av\\\\t =\frac{V}{Av}

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = \frac{\pi (0.015875\ m)^2}{4} = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [\frac{\pi (9.144\ m)^2}{4}][1.524\ m] = 100.1 m³

Therefore,

t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0
3 years ago
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