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garik1379 [7]
4 years ago
8

Which of the following is false?

Physics
1 answer:
stealth61 [152]4 years ago
3 0
The correct answer is C
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A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change. A) True B) False
bixtya [17]

Answer:

False

Explanation:

Faraday's law gives the relationship between the induced emf and the rate of change of magnetic flux i.e.

\epsilon=\dfrac{-d\phi}{dt}

The given statement "A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change" is false. The reason is that if the rate of change of magnetic flux is greater, then its will induce more emf. It would mean it does not say about emf.

Hence, it is false.

3 0
4 years ago
You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0
3 years ago
Which example demonstrates constant speed with changing direction?
Darya [45]
Ffffffddddffffffffffc
3 0
3 years ago
A gas is compressed from an initial volume of 5.75 L to a final volume of 1.23 L by an external pressure of 1.00 atm. During the
nataly862011 [7]

Answer:

<em>The internal energy change is 330.01 J</em>

Explanation:

Given

v_{1} the initial volume = 5.75 L

v_{2} the final volume = 1.23 L

P_{e} is the external pressure = 1.00 atm

q the heat energy removed = -128 J (since is removed from the system)

expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;

W = -P_{e}ΔV

W = -1.00 x(1.23 - 5.75)

W =  -1.00 x -4.52

W = 4.52 L atm

converting to joules we have

W = 4.52 L atm x 101.33 J/ L atm = 458.01 J

The internal energy change during compression can be calculated thus;

ΔU = q + W

ΔU = -128 J + 458.01 J

ΔU = 330.01 J

Therefore the internal energy change is 330.01 J

8 0
3 years ago
When mantle rocks near the radioactive core are heated, they become less dense than the cooler, upper mantle rocks. These warmer
avanturin [10]
C.convection cells is the right answer
4 0
3 years ago
Read 2 more answers
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