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3 years ago

According to Newton's 3rd law of motion, an action creates _____.

Physics

2 answers:

Katena32 [7]3 years ago

7 0

it's (c) an equal and opposite reaction

miv72 [106K]3 years ago

5 0

C.an equal and opposite reaction

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A satellite moves in a circular orbit a distance of 1.6×10^5 m above Earth's surface. (radius of Earth is 6.38 x 10^6m and its m

Rashid [163]

Answer:

The speed of the satellite is 7809.52 m/s

Explanation:

It is given that,

Radius of Earth, $r=6.38\times 10^6\ m$

Mass of earth, $M=5.98\times 10^{24}\ kg$

A satellite moves in a circular orbit a distance of, $d=1.6\times 10^5\ m$ above Earth's surface.

We need to find the speed of the satellite. It is given by :

$v=\sqrt{\dfrac{GM}{R}}$

R = r + d

$R=(6.38\times 10^6\ m+1.6\times 10^5\ m)=6540000\ m$

So, $v=\sqrt{\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 5.98\times 10^{24}\ kg}{6540000\ m}}$

v = 7809.52 m/s

So, the speed of the satellite is 7809.52 m/s. Hence, this is the required solution.

5 0

3 years ago

30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

schepotkina [342]

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

$P=\frac{W}{T}$

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

$P=\frac{W}{T}$

Later, the force does six times more work, so the work done now is

$W'=6W$

And this work is done in half the time, so the new time is

$T'=\frac{T}{2}$

Substituting into the equation of the power, we find the new power produced:

$P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P$

So, 12 times more power.

4 0

3 years ago

How do the different forms of potential energy depend on an object's position or chemical composition?

devlian [24]

It depends on the objects chemical composition.

6 0

2 years ago

Determine the kinetic energy of a 2000 kg roller coaster car that is moving at the speed of 10 ms

kondaur [170]

Answer:

$\boxed {\boxed {\sf 100,000 \ Joules}}$

Explanation:

Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

$E_k= \frac{1}{2} mv^2$

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.

m= 2000 kg
s= 10 m/s

Substitute these values into the formula.

$E_k= \frac{1}{2} (2000 \ kg ) \times (10 \ m/s)^2$

Solve the exponent.

(10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

$E_k= \frac{1}{2} (2000 \ kg ) \times (100 \ m^2/s^2)$

Multiply the first two numbers together.

$E_k= 1000 \ kg \times (100 \ m^2/s^2)$

Multiply again.

$E_k= 100,000 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

$E_k= 100,000 \ J$

The roller coaster car has <u>100,000 Joules</u> of kinetic energy.

3 0

3 years ago

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago