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AlekseyPX
3 years ago
11

Explain why wedges and screws are actually types of inclined planes.

Physics
2 answers:
slava [35]3 years ago
5 0
<span>The screw is really just an inclined plane covered around with a tiny pole. The wedge is definitely an inclined plane, since it starts with a point, then rises getting thicker, as an inclined plane. </span>
oee [108]3 years ago
4 0
The screw is really an inclined plane cover around with a tiny pole. The wedge is definitely an inclined plane, sinse it starts with  point, then rises getting thicker, as on inclined plane.
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Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 25,000 J as the
salantis [7]

Answer:

568.18 N

Explanation:

From the question,

The formula for gravitational potential is given as

Ep = mgh........................ Equation 1

Where Ep = Gravitational potential, m = mass of the diver,h = Height.

But,

W = mg.................... Equation 2

Where W = weight of the diver.

Substitute equation 2 into equation 1

Ep = Wh

Make W the subject of the equation

W = Ep/h................... Equation 3

Given: Ep = 25000 J, h = 44 m

Substitute into equation 3

W = 25000/44

W = 568.18 N.

Hence the weight of the diver = 568.18 N

5 0
3 years ago
If a receiver is overly selective:
VMariaS [17]

Answer:

C) only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results.

Explanation:

Selectivity is the ability of a receiver to respond only to a specific signal on a wanted frequency and reject other signals nearby in frequency.

If a receiver is overly selective, only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. Whereas, if a receiver is underselective, the receiver can pick different signals on different frequencies at the same time.

7 0
2 years ago
Let's say that we have a pilot that is dropping a package from a plane that is flying horizontally at a constant speed. If we ne
antoniya [11.8K]

Answer:

The package will be directly below the location of the plane.

Explanation:

Look up projectile motion for more information. The horizontal speed of the package is separate from the vertical speed of the package. The vertical speed of the falling package will be based on the rate of acceleration and the height of the package when dropped. The horizontal speed of the package will be the same as the plane so the package will remain directly below the plane the entire time until the package hits the ground.

6 0
2 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
A student puts a strip of litmus paper into a solution and the paper becomes red in color. What does this tell you about the sol
natita [175]
Litmus paper turns red under acidic conditions. Hope this helped!
3 0
3 years ago
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