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Leto [7]
2 years ago
8

2. What is the total effect of sound produced in an enclosed space called?

Physics
1 answer:
Igoryamba2 years ago
6 0
I believe it is Acoustics, the real name for it would be like Reverberation. But go with Acoustics. :)
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(Thought question) At noon On June 21 will the shadow length of 42 degrees north latitude be shorter or longer than it is on Mar
g100num [7]

Answer: Shorter

Explanation: Shadow is formed when an light source is obstructed by an opaque object. The closer the source, shorter is the length of the shadow. In fact, when the source is exactly overhead, no shadow of the object is formed.

June 21 marks the Summer solstice which means the Sun passes directly overhead Tropic of cancer (23.5° N) at noon. March 21 marks the equinox which means sun passes directly overhead equator (0°).

Shadow length of an object at 42° Northern latitude will be shorter on June 21 because the Sun will be closer to this latitude as compared to March 21.

5 0
3 years ago
What is the connection between the liver (organ), UV radiation (sunlight), and bone tissue
Digiron [165]
The answer to this question would be: vitamin D

The UV is needed by the skin to make previtamin D3. Previtamin D3 or cholecalciferol made from the skin will be changed in the liver into 25- hydroxyvitamin D3 and then sent to the kidney to be changed into 1,25- dihydroxyvitamin D3. The last change in the kidney will active the vitamin D. Vitamin D has a role in the calcium absorption which was will strengthen the bone tissue.
6 0
3 years ago
A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha
Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss} (1)

From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that m = 175\,kg, g = 9.807\,\frac{m}{s^{2}}, h_{1} = 18\,m, h_{2} = 8\,m and v_{2} = 11\,\frac{m}{s}, then the work done by non-conservative force is:

W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0
2 years ago
A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v
kifflom [539]

Here we will say that there is no external torque on the system so we will have

L_i = L_f

here we know that

L_i = I_1\omega_1

where we know that

I_1 = \frac{2}{5}mr^2

Also we know that

I_2 = \frac{1}{12}mL^2

initial angular speed will be

\omega_1 = 2\pi(2rev/s) = 4\pi rad/s

now from above equation

\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega

0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega

0.452 = 0.1875 \omega

now we have

\omega = 2.41 rad/s

so final speed will be 2.41 rad/s

6 0
3 years ago
A television set is plugged into a 120 V outlet. The television circuit carries a current equal to 0.75 A. What is the overall r
Arlecino [84]

Answer:

20,00

Explanation:

This is the answer

7 0
2 years ago
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