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guajiro [1.7K]
3 years ago
11

What happens to the particles of matter vibrated by a wave after the wave passes through the matter

Physics
1 answer:
Mariulka [41]3 years ago
5 0

Answer:

As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on.

Explanation:

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A 16 lb block rests on a horizontal frictionless surface. A cord attached to the block, running horizontally, passes over a pull
quester [9]

Answer:

I = 6.2161900319309  slug-feet^2[/tex]

Explanation:

The moment of inertia of an object may simply be stated as a measure of how difficult it is to start it spinning, or to alter an object's spinning motion

The moment of inertia of an object is given by the expression

I=mr^{2}

but in this case we are talking about the moment of inertia of a circular disk, the expression is a bit different

I=[tex]\frac{1}{2}mr^{2}[/tex]

inputting the values given in the expression above

I=\frac{1}{2}(16)5^{2}

moment of inertia  = 200 lb

the expected outcome should be in slug feet squared

1 slug-feet squared = 32.1740488782426

therefore we need to divide our answer by this value

so

= \frac{200}{32.1740488782426} \\= 6.2161900319309  slug-feet^2

3 0
3 years ago
A person gets a running start toward a wall from 5 meters away. The person pushes the wall with a force of 500 newtons, but the
Ostrovityanka [42]
100 joules is the answer.
6 0
3 years ago
Read 2 more answers
A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a hori
marshall27 [118]

Answer:

K_2=6.4J

Explanation:

According to the principle of conservation of momentum, we have:

\Delta p=0\\p_i=p_f\\m_1v_1_i +m_2v_2_i=m_1v_1_f+m_2v_2_f

Here 1 is for the bullet and 2 is for the block. Since the block is initially at rest v_i_2=0. Solving for v_2_f and replacing the given values:

v_2_f=\frac{m_1(v_1_i-v_1_f)}{m_2}\\v_2_f=\frac{10*10^{-3}kg(1.8\frac{km}{s}-1\frac{km}{s})}{5kg}\\v_2_f=0.0016\frac{km}{s}*\frac{1000m}{1km}=1.6\frac{m}{s}

The kinetic energy of the block is given by:

K_2=\frac{m_2(v_f_2)^2}{2}\\K_2=\frac{5kg(1.6\frac{m}{s})^2}{2}\\K_2=6.4J

3 0
3 years ago
3. Mr. Hogan in his 3000 kg truck is driving down
S_A_V [24]

Answer:

|I|=21,000\ kg.m/s

<em>Correct answer: Option E</em>

Explanation:

<u>Impulse and Momentum </u>

The Impulse is defined as the change of momentum of an object in a certain interval. The formula is:

I=\Delta p=p_f-p_o

Where pf and po are the final and initial momentums respectively. Knowing that for m= mass of the object and v=velocity (or speed in the scalar form)

p=m.v

Thus

I=m(v_f-v_o)

Mr. Hogan's 3000 kg truck is initially at 36 m/s and later slows down to 29 m/s. We compute the impulse as follows

I=3000\ kg(29\ m/s-36\ m/s)

I=3000(-7)\ kg.m/s

I=-21,000\ kg.m/s

We are asked for the magnitude, thus

|I|=21,000\ kg.m/s

Correct Answer: Option E

6 0
3 years ago
As viewed from above, a swimming pool has the shape of the ellipse given by x21600+y2400=1 The cross sections perpendicular to t
Law Incorporation [45]

Answer:

the total volume of the pool = 85,333.3ft^3

Explanation:

Since the cross sections perpendicular to the ground and parallel to the y-axis are squares.This means that the dimensions of each square-shaped slice are:

width = 2y

depth = 2y

where y is a function of x.

The thickness of each slice is equal to dx

The area of each cross-sectional slice is

(2y)(2y) = 4y^2

The volume of each cross-section is

dV = 4y^2 dx. .....1

Whe need to write y as a function of x.

This comes from the equation of the ellipse:

x^2/a^2+y^2/b^2=1

with a^2 = 1600 and b^2 = 400

Then a = 40 and b = 20

Rearranging the equation gives:

y^2 = b^2(1 - x^2/a^2)

This can be substituted into the equation for dV above: (eqn 1 above.)

dV = 4b^2(1 - x^2/a^2)dx

The volume of the pool is the sum of all of the slices, which you calculate by integrating dV. However, due to symmetry, you can calculate the volume of half of the pool, and then multiply by 2 to get the total volume. This means that you integrate from the origin of the axes (x=0) to the edge of the pool (x=a).

We will denote the “half-volume” by Vh.

Vh= integral of dV from 0 to a

Vh = [4b^2( x - x^3/3a^2) ] from 0 to a

Substituting a we have.

Vh = 4b^2(a - a/3)

Vh = (8/3)ab^2

V = 2Vh = (16/3)ab^2

Substituting the values of a and b

V = (16/3)(40 × 20^2)

V = 85,333.3ft^3

7 0
3 years ago
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