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skad [1K]
3 years ago
8

Give at least 4 human activities which demonstrate the involvement of inertial frame of reference​

Physics
1 answer:
Sergio039 [100]3 years ago
7 0
One clearcut example of an inertial reference frame is an isolated spaceship, far, far away from the Earth, the Sun, the Milky Way Galaxy, and all other massive objects. Fred places a blue ball into a claw at the left end of the ship, and red ball into a claw at the right end of the ship.
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How does the direction of the electric current, moving across a battery powered device, differ from the direction of travel in t
svetoff [14.1K]
They both have different wave traction's
8 0
3 years ago
Read 2 more answers
Which of the following options is correct and why?
Dimas [21]

Answer:

Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.

Explanation:

So, we are given the following set of infomation in the question given above;

=> "spherical Gaussian surface of radius R centered at the origin."

=> " A charge Q is placed inside the sphere."

So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?

The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;

REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.

Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.

6 0
3 years ago
Question #6a) You were told to assume that the ball bounced to a height of 1.2 m and that the distance between the bounces measu
Inessa [10]

Answer:

Explanation:

In the problem they give the case of a ball that bounces on the vertical axis the height is getting smaller, on the x axis it is the same distance all the time.

Questions about the aspect of the x-axis

- Since the distance traveled between the rebounds is the same, the speed must be the same or constant throughout the entire journey.

- Since the distance and speed are equal, the time between rebounds is the same

- There can be no acceleration because the bounce gap should change, the acceleration is zero all the time

Questions about the Y axis

- The vertical speed of the first boat would be the greatest of all, so it has the highest height

b) Evidence or not Evidence

1  evidence, because the graphs are different, one is a straight line and he gives a parable

2 no evidence, nothing involved on the x axis

3  no evidence, the acceleration on the y axis is independent of the acceleration on the x axis

4  no evidence, the time that is cast is the only change that is the same for both movements

5  evidence, the fact that no mixture of components is found allows the variable to be separated into different equations

6  no evidence, says nothing about the x-axis

4 0
3 years ago
According to the Newton’s first law of motion , which of the following objects will change in motion
Kobotan [32]
The correct answer is C
4 0
3 years ago
Calculate the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of
vesna_86 [32]

Answer:

0.21%

Explanation:

We are given;

Mass; m = 100 kg

Diameter; d = 2.2 mm = 2.2 × 10^(-3) m

Young's modulus; E = 12.5 x 10^(10) N/m².

Formula for area is;

A = πd²/4

A = (π/4) x (2.2 x 10^(-3))²

A = 3.8 x 10^(-6) m²

Force; F = mg

g is acceleration due to gravity and has a constant value of 9.8 m/s²

F = 100 × 9.8

F = 980 N

Formula for young's modulus is;

E = Stress/strain

Formula for stress = F/A

Formula for strain = ΔL/L

Thus;

E = (F/A)/(ΔL/L)

Making ΔL/L the subject, we have;

ΔL/L = (F/A)/E

Plugging in the relevant values;

ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))

ΔL/L = 0.0021

Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%

8 0
3 years ago
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