How are waves that are created by a musical instrument and waves that

A shear force of 400 N is applied to one face of an aluminum cube with sides of 30 cm. What is the resulting relative displaceme

shtirl [24]

Answer:

5.3 × 10^(-8) m

Explanation:

We are given;

Shear force; F = 400 N

Length of cube; L_o = 30 cm = 0.3 m

Shear modulus; S = 2.5 × 10^(10) N/m²

Now,the resulting relative displacement can be gotten from the formula;

F = A × S × Δx/L_o

Where Δx is resulting relative displacement

A is area.

Area of cube = (L_o)² = 0.3² = 0.09

Thus, making Δx the subject, we have;

Δx = (F × L_o)/(A × S)

Plugging in the relevant values;

Δx = (400 × 0.3)/(0.09 × 2.5 × 10^(10))

Δx = 5.3 × 10^(-8) m

6 0

3 years ago

A cannonball with a mass of 8.3 kg is fired from a cannon that has a mass of 949.3 kg. The cannonball is fired with a speed of 1

Fittoniya [83]

conservation of momentum
mass(ball) × speed(ball) = mass(cannon) × speed(cannon)
(8.3 kg) × (187.9 m/s) = (949.3 kg) × v
1.64 m/s = v

6 0

3 years ago

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = 0.11 J

3 0

3 years ago

Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL << L). Demuestre que su período aumenta en: ΔT = π

Kruka [31]

Answer:

ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

w = $\sqrt{\frac{g}{L} }$

angular velocity and period are related

w = 2π / T

we substitute

2π / T = \sqrt{\frac{g}{L} }

T = $2\pi \ \sqrt{\frac{L}{g} }$

In this exercise indicate that for a long Lo the period is To, then and increase the long

L = L₀ + ΔL

we substitute

T = $2\pi \ \sqrt{\frac{L + \Delta L}{g} }$

T = $2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }$

in general the length increments are small ΔL/L «1, let's use a series expansion

$\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...$

we keep the linear term, let's substitute

T = $2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )$

if we do

T = T₀ + ΔT

T₀ + ΔT = $2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}$

T₀ + ΔT = T₀ + $\pi \sqrt{\frac{1}{Lg} } \ \Delta L$

ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

4 0

3 years ago

A trout was swimming in a river. The trout swam 7,000centimeters directly upstream at a constant velocity. It swam that distance

Mars2501 [29]

Answer:

10000

Explanation:

7 0

3 years ago

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