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Ugo [173]
2 years ago
15

A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density

of water is 62.4 lb/ft3).
Physics
1 answer:
blsea [12.9K]2 years ago
6 0

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}

F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

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A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t
Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

3 0
2 years ago
What must be true about a surface in order for diffuse reflection to occur?
balu736 [363]

Answer:

carpet

Explanation:

Diffuse reflection is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection.

The structure of carpet's surface is as shown. Thus it shows large amount of diffuse reflection.

4 0
3 years ago
Whats the force of gravitation of a 10kg rock and 100kg boulder which are 5 meters apart​
spayn [35]

Answer:

F = 2.6692 x 10⁻⁹ N

Explanation:

Given,

The mass of the rock, m = 10 kg

The mass of the boulder, M = 100 kg

The distance between them, d = 5 m

The gravitational force between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by the formula

                                   <em> F = GMm/d²  newton</em>

Where,

                                 G - Universal gravitational constant

Substituting the given values,

                                 F = 6.673 x 10⁻¹¹ x 100 x 10 / 5²

                                 F = 2.6692 X 10⁻⁹ N

Hence, the force between the two bodies is, F = 2.6692 X 10⁻⁹ N

6 0
3 years ago
As an object falls freely toward the earth, the momentum of the object-earth system (1) decreases (2) increases (3) remains the
Brums [2.3K]

The best and most correct answer among the choices provided by your question is the third choice or number 3.

<span>As an object falls freely toward the earth, the momentum of the object-earth system remains the same.</span>


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
4 0
3 years ago
A car moves around a circular track at a constant rate. What must change?
suter [353]

B. only its velocity should change

4 0
2 years ago
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