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lidiya [134]
2 years ago
5

The label has been scratched off a tuning fork and you need to know its frequency. From its size, you suspect that it is somewhe

re around 250 Hz. You find a 250-Hz tuning fork and a 270-Hz tuning fork. When you strike the 250-Hz fork and the fork of unknown frequency, you hear a modulation (fluctuation of the amplitude) at a frequency of 5 Hz. When you strike the unknown with the 270-Hz fork, you hear a modulation at a frequency of 15 Hz. What is the unknown frequency
Physics
1 answer:
bonufazy [111]2 years ago
3 0

Answer:

255 Hz

Explanation:

With 5 beats per second with the 250 Hz fork, we know the unknown fork is either 250 - 5 = 245Hz or 250 + 5 = 255 Hz

With 15 beats per second with the 270 Hz fork, we know the unknown fork is either 270 - 15 = 255Hz or 270 + 15 = 285 Hz (most people would have a hard time discerning 15 beats per second... 5 per second is hard enough)

As 255 is the common frequency, it is the one selected.

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Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M
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The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

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M_1 g - T = M_1 a

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a is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, M_2 g sin \theta. So the equation of motion is

T-M_2 g sin \theta = M_2 a

where

M_2 = 13.5 kg is the mass of the 2nd block

\theta=35.5^{\circ} is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

a=0

So the two equations become:

M_1 g - T=0\\T-M_2 g sin \theta = 0

Isolating T from the 1st equation,

T=M_1 g

And substituting into the 2nd equation, we can find the value of the mass M_1:

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7 0
3 years ago
How much ice (at 0°C) must be added to 1.90 kg of water at 79 °C so as to end up with all liquid at 8 °C? (ci = 2000 J/(kg.°
muminat

m = mass of the ice added = ?

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using conservation of heat

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inserting the values

m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)

m = 1.53 kg

4 0
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