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Setler [38]
3 years ago
8

If a wave has a frequency of 50 Hz and a wavelength of 3 meters, what is its Velocity? (Use

Physics
1 answer:
allochka39001 [22]3 years ago
4 0

Answer:

the answer is 150m/s

Explanation:

v=λf

v=50*3

v=150m/s

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What is the area of the parallelogram?<br>square units​
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The area is the length times the height so A= bh ...?
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Why do you see objects when you shine a flashlight in a dark room?​
Afina-wow [57]

Answer:

We see objects in a dark room due to the emission of light photons which are sensitive to our eyes. Darkness is simply a terminology used to describe the absence of light. Visible light to human is a component of the electromagnetic spectrum. Our eyes have receptors that picks the photons which light releases

Explanation:

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3 years ago
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The latent heat of fusion of alcohol is 25 kcal/kg and its melting point is -114 o C. It has a specific heat of 0.60 in its liqu
Elanso [62]

Answer:

Total energy is 170 kJ

Explanation:

Given data:

latent heat of fusion of alcohol is 25 kcal/kg

melting point of alcohol is -114 degree c

specific heat us 0.60 k cal/kg degree c

energy need for 2 kg solid alcohol is

for Melting:

Energy Q is calculated as

Energy, Q = 25 \times 2.0 kg = 50 kJ

Energy required for Heating liquid:

Energy, ΔH = 2.0 kg \times 0.60 \times (100°C) = 120 kJ

Total  energy = (50 kJ + 120 kJ) = 170 kJ

4 0
3 years ago
The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Dril
STatiana [176]

Answer:

The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

Explanation:

Given:-

- The diameter of the drill bit, d = 98 cm

- The power at which drill works, P = 5.85 hp

- The rotational speed of drill, N = 1900 rpm

Find:-

What Torque And Force Is Applied To The Drill Bit?

Solution:-

- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).

- The relation between these quantities is given:

                         T = 5252*P / N

                         T = 5252*5.85 / 1900

                         T = 16.171 Nm

- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):

                          T = F*r

Where,   r = d / 2

                          F = 2T / d

                          F = 2*16.171 / 0.98

                          F = 33 N

Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

4 0
3 years ago
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A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
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