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leonid [27]
2 years ago
6

A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from i

t at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Physics
1 answer:
murzikaleks [220]2 years ago
4 0

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

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Bromine is a liquid at room temperature. The volume of a sample of bromine is measured in a 50 beaker and a 100 ml beaker. How w
Stells [14]

Answer:

The comparisons are;

The height of the bromine in the 50 ml beaker will be twice that of the 100 ml beaker

The measurement of the volume with the 50 ml beaker will be more accurate than the measurement taken with the 100 ml beaker, because the differences in the height of the bromine in the 50 ml beaker is more obvious than the differences measured with the 100 ml beaker.

The actual volume of bromine in both beakers will be equivalent

Explanation:

The properties of a liquid are;

1) The volume of a liquid is relatively fixed at conditions that are suitable for it to remain in the liquid state compared to the volume occupied by a gas

2) A liquid will assume the shape of a container in which it is placed

3) The surface of a liquid in a container is flat due in order that the attractive forces between the molecules of the liquid at the surface and inside the body of the liquid should be in equilibrium

Therefore, given that the volume of the Bromine is measured in 50 ml beaker and a 100 ml beaker, there will be differences in the measured height of the same volume of bromine in each beaker.

5 0
3 years ago
What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged partic
VladimirAG [237]

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

A

   v_w  =  2.256 *10^{8} \  m/s

B

  v_e  =  2.21 *10^{8} \  m/s

C

The  correct option is  B  

Explanation:

From the question we are told that

      The refractive  index of water is  n_w  =  1.33

      The  refractive  index of ethanol is  n_e  =  1.36

       

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

       v_w  =  \frac{c}{n_w }

Where  c is the speed of light with value  c =  3.0 *10^{8} \  m/s

       v_w  =  \frac{3.0 *10^{8}}{1.33 }

       v_w  =  2.256 *10^{8} \  m/s

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

            v_e  =  \frac{ c}{n_e }

  =>       v_e  =  \frac{3.0 *10^{8}}{1.36 }

=>          v_e  =  2.21 *10^{8} \  m/s

4 0
3 years ago
Click on the boxes below to indicate the number of electrons or protons in each labeled position for a zinc atom (z=30, a=65). P
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Answer:

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Symbol: Zn

Atomic Number: 30

Atomic Mass: 65.39 amu

Melting Point: 419.58 °C (692.73 K, 787.24396 °F)

Boiling Point: 907.0 °C (1180.15 K, 1664.6 °F)

Number of Protons/Electrons: 30

Number of Neutrons: 35

Classification: Transition metal

Crystal Structure: Hexagonal

Density at 293 K: 7.133 g/cm3

Color: bluish

('lil long, sorry)

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I think the answer would be Car C.
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A large force is required to accelerate the mass of the bicycle and rider. Once the desired constant velocity is reached, a much smaller force is sufficient to overcome the ever-present frictional forces. 
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