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klemol [59]
3 years ago
14

1. What could scientists learn by using the Law of Superposition?

Physics
2 answers:
Juli2301 [7.4K]3 years ago
7 0
Answer C. 1.scientists learn by using the law of superposition B
natta225 [31]3 years ago
3 0
You answer would be C
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Which of the following refers to the ability to do work
valina [46]
The ability to do work is called energy.
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4 years ago
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A particle at 9 AM is moving towards the east at 4 ms At 12 noon, it changes its velocity and starts moving towards the north un
nexus9112 [7]

Answer:

Explanation:

Acceleration is the time rate of change of velocity.

Acceleration and velocity are vectors

If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.

There are 3 hours or 10800 seconds between 10 AM and 1 PM

a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²

There are 14400 seconds between 10 AM and 2 PM

The velocity changes are still the same

a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²

7 0
3 years ago
How do you solve this scientific notation question?
podryga [215]

factorise 1/10^2 out of both.

add 1/4.5 to 1/9.4 .... common denominator  4.5x9.4

multiply result by 1/10^2

find reciprocal of result

phew !

8 0
3 years ago
Once out in space, the space shuttle needs to _____ to keep going in the same direction at the same speed.
erik [133]

Answer:

C

Explanation:

because it not running will set it off course but running all the time it will run. out of fuel

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The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser
Dafna1 [17]

A) Orbital speed: v=\sqrt{\frac{GM}{R}}

B) Kinetic energy: K= \frac{GmM}{2R}

D) The orbital period is T=\frac{2\pi}{\sqrt{GM}}R^{3/2}

F) The angular momentum is L=m\sqrt{GMR}

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

A)

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

F=G\frac{mM}{R^2}

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

F=m\frac{v^2}{R}

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

G\frac{mM}{R^2}=m\frac{v^2}{r}

And solving for v, we find

v=\sqrt{\frac{GM}{R}}

B)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

v=\sqrt{\frac{GM}{R}} is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}

D)

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

v=\frac{2\pi R}{T}

where

2\pi R is the circumference of the orbit

T is the orbital period

We already found that the orbital speed is

v=\sqrt{\frac{GM}{R}}

Substituting into the equation,

\sqrt{\frac{GM}{R}}=\frac{2\pi R}{T}

And making T the subject,

T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}}=\frac{2\pi}{\sqrt{GM}}R^{3/2}

F)

The angular momentum of an object is defined as

L=mvr

where

m is the mass of the object

v is its speed

r is the radius of the orbit

For the satellite here we have

m (mass of the satellite)

v=\sqrt{\frac{GM}{R}} (orbital speed)

R (orbital radius)

Substituting,

L=m\sqrt{\frac{GM}{R}}R=m\sqrt{GMR}

G)

First, we rewrite the list of expressions for the different quantities that we found:

Orbital speed: v=\sqrt{\frac{GM}{R}}

Kinetic energy: K= \frac{GmM}{2R}

Orbital period: T=\frac{2\pi}{\sqrt{GM}}R^{3/2}

Angular momentum: L=m\sqrt{GMR}

Now we observed the dependence of each quantity from R:

Orbital speed: v\propto R^{-1/2}

Kinetic energy: K \propto R^{-1}

Orbital period: T \propto R^{3/2}

Angular momentum: L \propto R^{1/2}

So the exponent of the radial dependence of each quantity is:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0
3 years ago
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