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timofeeve [1]
3 years ago
11

Answer all these questions

Physics
2 answers:
Mama L [17]3 years ago
6 0

Answer:

1 . Soundwave

2. pitch are sounds that are high or low corresponding and according to a certain frequency measured in Hertz

3. volume is the increase in sound or decrease in sound how loud or soft something is while amplitude is strength of sound

4. frequency is measured in Hertz one wave per second and wavelength is distance between corresponding points of two recurring and consecutive sound waves

Kazeer [188]3 years ago
5 0

Answer:

mechanical waves,

.

the quality of a sound governed by the rate of vibrations producing it; the degree of highness or lowness of a tone.

.

If the amplitude increases the volume increases and vice versa.

.

The type of medium affects a sound wave as sound travels with the help of the vibration in particles.

.

The higher the frequency, the shorter the wavelength.

Explanation:

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The correct answer is D. I alread took this test.

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Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer
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Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0
3 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

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We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

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A diagram of a closed circuit with power source on the left labeled 30 V, a resistor on the top labeled 2 Ohms, a resistor on th
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