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Bess [88]
3 years ago
5

A motor is used to produce

Physics
1 answer:
Tpy6a [65]3 years ago
4 0
Your answer would be 4.0 hz.
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What do echolocation and ultrasounds have in common?
Natali5045456 [20]
They both are mechanical waves.

Explanation:

Echolocation and ultrasounds are both mechanical waves that require a medium to trace through. They both process and transfer information through waves; the difference is that we humans cannot hear ultrasound waves.
8 0
2 years ago
PLS HELP WILL MARK BRAINLIEST IF RIGHT NEED IMMEDIATELY PLEASEEEEE
timofeeve [1]

Answer:

Explanation:

Prescribe medications.

Prescribe treatments or therapies.

Treat patients using psychological therapies.

Collect medical information from patients, family members, or other medical professionals.

Record patient medical histories.

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3 0
2 years ago
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What is the weight of the ethyl alcohol that exactly fills a 200.0 mL container? The density of ethyl alcohol is 0.789 g/mL
sweet [91]

Answer:

1.55 N

Explanation:

Density = mass / volume

0.789 g/mL = m / 200.0 mL

m = 157.8 g

Weight = mass × acceleration due to gravity

W = (0.1578 kg) (9.8 m/s²)

W = 1.55 N

3 0
3 years ago
g We saw in class that in a pendulum the string does no work. We also saw that the normal force does no work on an object slidin
Ymorist [56]

Answer:

From question (a) and (b) the pendulum motion is perpendicular to the force so the normal force will do no work and the tension in the string of the pendulum will not work

       i.e Normal \ Force(N)  = mg cos \theta

And \theta = 90 so

           N = 0

c

An example will be a where a stone is attached to the end of a string and is made to move in a circular motion while keeping the other end of the string in a fixed position        

d

A dog walking along a surface which has friction, here the frictional force would acting in the direction of the motion and this would do positive work  

Explanation:

5 0
3 years ago
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To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz ,
bazaltina [42]

Answer:

Length of the pipe = 53.125 cm

Explanation:

given data

harmonic frequency f1  = 800 Hz

harmonic frequency f2  = 1120 Hz

harmonic frequency f3  = 1440 Hz

solution

first we get here fundamental frequency that  is express as

2F = f2 - f1    ...............1

put here value

2F = 1120 - 800

F = 160 Hz

and

Wavelength is express as

Wavelength  = Speed ÷ Fundamental frequency    ................2

here speed of waves in air  = 340 m/s

so put here value

Wavelength  =340 ÷ 160

Wavelength   = 2.125 m

so

Length of the pipe will be

Length of the pipe = 0.25 × wavelength    ......................3

put here value

Length of the pipe = 0.25 × 2.125

Length of the pipe = 0.53125 m

Length of the pipe = 53.125 cm

7 0
2 years ago
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