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Usimov [2.4K]
3 years ago
13

State four components of a circuit

Physics
1 answer:
makvit [3.9K]3 years ago
7 0

Answer:

» Battery or cell

» Switch

» Connecting wires

» Resistor or appliance

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Find the first three harmonics of a string of linear mass density 2. 00 g/m and length 0. 600 m when the tension in it is 50. 0
gavmur [86]

The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

<h3>Velocity of the wave</h3>

The velocity of the wave is calculated as follows;

v = √T/μ

where;

  • T is tension
  • μ is mass per unit length = 2 g/m = 0.002 kg/m

v = √(50/0.002)

v = 158.1 m/s

<h3>First harmonic or fundamental frequency of the wave</h3>

f₀ = v/λ

where;

  • λ is the wavelength = 2L

f₀ = v/2L

f₀ = 158.1/(2 x 0.6)

f₀ = 131.8 Hz

<h3>Second harmonic of the wave</h3>

f₁ = 2f₀

f₁ = 2(131.8 Hz)

f₁ = 263.6 Hz

<h3>Third harmonic of the wave</h3>

f₂ = 3f₀

f₂ = 3(131.8 Hz)

f₂ = 395.4 Hz

Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

Learn more about harmonics here: brainly.com/question/4290297

#SPJ1

6 0
1 year ago
How can the difference in the brightness of spectral lines be explained?
Ksenya-84 [330]
<span>more lines = a lot of electrons returning back to ground state from same level</span>
5 0
3 years ago
A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?
Rashid [163]

Answer:

<h2>3 m/s^2</h2>

Explanation:

Step one:

given

Mass m= 4kg

Force F= 12N

Required

Acceleration the relation between force, acceleration, and mass is Newton's first equation of motion, which says a body will continue to be at rest or uniform motion unless acted upon by an external force

F=ma

a=F/m

a=12/4

a=3 m/s^2

4 0
2 years ago
Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s
Zielflug [23.3K]

Answer:

there will be collision

Explanation:

v_{s} =  speed of sue = 34 m/s

v_{v} = speed of van = 5.20 m/s

v_{sv} = speed of sue relative to van  = v_{s} - v_{v} = 34 - 5.20 = 28.8 m/s

d_{s} = stopping distance after brakes are applied

D = distance between sue and van = 160 m

v_{f} = final speed of sue = 0 m/s

a = acceleration = - 1.80 m/s²

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a d_{s}

0^{2} = 28.8^{2} + 2 (1.80) d_{s}

d_{s} = 230.4 m

Since  d_{s} < D

hence there will be collision

7 0
3 years ago
How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium
Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where;

  • V₁ = AL
  • V₂ = A(L - y)
  • P₁ = Pa
  • P₂ = Pa + ρgh
  • T₁ = 20⁰C = 293 K
  • T₂ = 10⁰ C = 283 k

\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\  +\  1000 \times  9.8 \times  100)} )\\\\y = 3.8 \ m

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

#SPJ4

3 0
2 years ago
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