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Serga [27]
3 years ago
5

what is the position of centre of curvature for concave and convex mirror show with the help of diagram if you can​

Physics
1 answer:
Anon25 [30]3 years ago
8 0

it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram

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At which angle must a laser beam enter the water for no refraction to occur?
abruzzese [7]

Light that enters the new medium <em>perpendicular to the surface</em> keeps sailing straight through the new medium unrefracted (in the same direction).

Perpendicular to the surface is the "normal" to the surface. So the angle of incidence (angle between the laser and the normal) is zero, and the law of refraction (just like the law of reflection) predicts an angle of zero between the normal and the refracted (or the reflected) beam.

Moral of the story:  If you want your laser to keep going in the same direction after it enters the water, or to bounce back in the same direction it came from when it hits the mirror, then shoot it <em>straight on</em> to the surface, perpendicular to it.

5 0
2 years ago
A box of paper is labeled 24lb paper. This means that 500 sheets (counted number) of paper size 17in x 22in weighs 24 pounds (Th
Tresset [83]

Answer:

The mass of a single paper  is approximately 0.047 lb/paper which in SI Units is approximately 21.77  g/paper

Explanation:

The given information on the size and the weight of paper are;

The mass of a box of 500 sheets of paper = 24 lb

The number of sheets in the paper = 500 sheets

The dimensions of the paper = 17 in. × 22 in., which is equivalent to  43.18 cm × 55.88 cm

The mass of a single paper = The mass of the box of paper/(The number of sheets of paper present in the box)

The mass of a single paper = 24 lb/500 = 0.047 lb/paper

Given that 1 lb = 453.6 g, we have;

0.047 lb/paper = 0.047 lb/paper×453.6 g/(lb) = 21.77  g/paper

The mass of a single paper  = 0.047 lb/paper = 21.77  g/paper.

6 0
2 years ago
When a car crashes, do you think its speed can make a difference in the<br> amount of damage done?
Oduvanchick [21]

Answer:

Yes

Explanation:

An increased speed will result in an increased amount of energy, so when it crashes some of that energy will bounce back and crumple the car.

3 0
3 years ago
Analyze data table #1 and then create a line graph comparing speed and time.
In-s [12.5K]
I don’t know what the answer is to the question but if I don’t answer the question I will be mad at the
8 0
3 years ago
The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte
cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

3 0
3 years ago
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