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babymother [125]
3 years ago
7

What is an inclined plane? How does it make our work easier?

Physics
1 answer:
Scorpion4ik [409]3 years ago
7 0
An inclined plane is a simple machine consisting of a sloping surface,used for raising heay bodies and it is easier to move an object.
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When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not n
Anarel [89]

Answer:

True, because unlike the apple we don't have a large as$ refrence point (the earth is too big to notice being pushed)

Explanation:

6 0
3 years ago
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A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m
Finger [1]

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

8 0
3 years ago
Heartburn is a form of indigestion felt as a burning sensation in the chest when stomach acid comes back up into the esophagus.
xenn [34]
The antacid is basic so it neutralizes acidity or lowers it. Then if it goes into the esophagus, it's not as strong and it doesn't hurt, and it also calms your stomach because the acidity in your stomach is also lower. Antacids are therefore taken by many people, especially as they grow older and things like heartburn become more common.
5 0
3 years ago
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The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of
PolarNik [594]
The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions
7 0
3 years ago
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A closed cylinder with a 0.15-m radius ends is in a uniform electric field of 300 n/c, perpendicular to the ends. the total flux
bixtya [17]

The total flux through the cylinder is zero.


In fact, the electric flux through a surface (for a uniform electric field) is given by:

\Phi = E A \cos \theta

where

E is the intensity of the electric field

A is the surface

\theta is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.


We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because \theta=90^{\circ} and \cos \theta=0).


On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

\Phi_1 = E \pi r^2

where r is the radius, and where we have taken \theta=0^{\circ} since the perpendicular to the surface is parallel to the direction of the electric field, so \cos \theta=1. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so \theta=180^{\circ} and \cos \theta=-1, therefore the flux is

\Phi_2 = -E \pi r^2

And the net flux through the cylinder is

\Phi = \Phi_1 + \Phi_2 = E \pi r^2 - E \pi r^2=0

4 0
3 years ago
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