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n200080 [17]
2 years ago
11

An organ pipe is tuned to exactly 384 Hz when the temperature in the room is 20°C. Later, when the air has warmed up to 25°C, th

e frequency is
A. Greater than 384 Hz.
B.384 Hz.
C. Less than 384 Hz.
Physics
1 answer:
maksim [4K]2 years ago
3 0

Answer: A. Greater than 384 Hz

Explanation:

The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.

Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.

Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

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The mechanical advantage is defined as the ratio between the force produced by a machine and the force applied in input:
MA= \frac{F_{out}}{F_{in}}
For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
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8, 8 W

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2 years ago
In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being
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Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun
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Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

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Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

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