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erastova [34]
3 years ago
10

You are looking at the image of a pencil in a mirror. What do you see in the mirror if the top half of the mirror is covered wit

h a piece of dark paper?
Physics
1 answer:
hoa [83]3 years ago
7 0

Answer:

Pencil's image will be visible.

Explanation:

Light is scattered off all the points of the pencil  and into all direction of space if light directed towards the mirror is reflected back into our eyes we can see the pencil.

If the top half of the mirror is covered with a piece of dark paper, Light is scattered off  the pencil  and reflected off the mirror can enter your eyes and you will see the image of the pencil.

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At an amusement park, a swimmer uses a water slide to enter the main pool. a. If the swimmer starts at rest, slides with negligi
adoni [48]

Answer:

a)6.7m/S

b)6.8m/s

Explanation:

Hello ! To solve the point b you must follow the steps below

1.Draw the slide taking into account its length and height and find the angle from which the swimmer is launched (see attached image)

2. Find the horizontal velocity (X) and vertical (Y) components (see attached image)

3) for the third step we must remember that as in the slide there is no horizontal acceleration the speed in X will remain constant at the end of the swimmer's path (Vx = 0.59m / s)

4)

the fourth step is to remember that vertically there is constant acceleration called gravity (g = 9.81m / s ^ 2), so to find the speed at the end of the route we use the following equation

Vfy= \sqrt{Vy^2+2gy}

where    

Vfy= final verticaly speed    

Vy=initial verticaly speed=0.59m/S

g=gravity=9.81m/S^2

y=height of slide=2.31m

solving

Vfy= \sqrt{Vy^2+2gy}\\Vfy= \sqrt{(0.59)^2+2(9.81)(2.31)}=6.77m/s

The last step is to add the velocity components vectorally at the end of the route with the following equation

V=\sqrt{Vfy^2+Vx^2} =\sqrt{6.77^2+0.59^2} =6.8m/s

point A

taking into account the previous steps we can infer that as the swimmer starts from rest, the velocity (Vx=Vy=O) is zero, so we should only use the formula for constant acceleration movement.

Vfy= \sqrt{Vy^2+2gy}

vy=0

Vfy=\sqrt{2gy}

Vfy=\sqrt{2(9.81)(2.31)}=6.7m/s

8 0
3 years ago
Calculate the impulse imparted when a 3,000-kg car hits a wall at 60 . m/s and comes to a stop.
Pani-rosa [81]

Impulse = change in momentum


The answer is 0.

7 0
3 years ago
ANSWER QUCIK<br> PLS THANK YOU
Reil [10]
Electrical energy in the charger and cable

Chemical energy in the battery of the mobile phone
8 0
3 years ago
A cat walks across the kitchen with a constant speed of 0.5 m/s. How long does it take the cat to walk 8 m?
ruslelena [56]
How does the cat walk-
4 0
3 years ago
A man pushes a 35.2 kg box across a frictionless floor with a force of 128 N. What is the acceleration of the box
vodka [1.7K]
The formula for Force is F = MA, or Force is equivalent to the product of Mass and Acceleration. 

F = 128N.
M = 35.2kg.

128 = 35.2A
Divide both sides by 35.2 to solve for the acceleration.
A = ~3.636

The acceleration is 3.636 m/s^2.

I hope this helps!
4 0
3 years ago
Read 2 more answers
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