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3 years ago

An output is a Push or pull _________________________ on the object

Physics

1 answer:

Ugo [173]3 years ago

5 0

Answer:

The input force that you use on an inclined plane is the force with which you push or pull an object. The output force is the force that you would need to lift the object without the inclined plane. This force is equal to the weight of the object.

Explanation:

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A farmer buys a piece of rectangular land with an area of 14 acres.He wishes to grow broccoli which requires 20" separation betw

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He now has no room left to hut checkens

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2 years ago

Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

$V=\frac {1}{3}\times \pi\times r^2\times h$

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

$V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h$

$V=\frac {550}{3549}\times h^3$

Also differentiating the expression of volume w.r.t. time as:

$\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}$

Given: $\frac {dV}{dt}$ = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

$-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}$

$\frac{55000}{1183}\times \frac {dh}{dt}=-4$

$\frac {dh}{dt}=-0.08603\ cm/s$

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

3 0

2 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

algol13

Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

<u>Explanation:</u>

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

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3 years ago

Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from a

riadik2000 [5.3K]

Answer:

Check the explanation

Explanation:

The beat frequency is

df = f2 - f1

the wavelength is

lamda1 = (v/f1)

and lamda2 = (v/f2)

where v = 340 m/s,f1 = 25.0 kHz and f2 = 20.0 kHz

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3 years ago

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