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katovenus [111]
3 years ago
10

Give two examples where friction is a nuisance

Physics
1 answer:
dlinn [17]3 years ago
8 0

Answer:

it can cause wear and tear and reduces efficiency as energy is lost.

Explanation:

You might be interested in
The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu
solniwko [45]

The frequency of the human ear canal is 2.92 kHz.

Explanation:

As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is  

Wavelength=4*Length of the ear canal

As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

wavelength = 4*2.9*10^{-2} =0.116

Then the frequency is determined as

f=c/λ=339/0.116=2922 Hz=2.92 kHz.

So, the frequency of the human ear canal is 2.92 kHz.

4 0
3 years ago
Sug<br>. To repeat the experiment what is it?<br>​
statuscvo [17]

Answer:

....more foam

Explanation:

7 0
3 years ago
A net force of 50 newtons is applied to a 20 kilogram cart that is already moving at 1 m/s the final speed of the cart was 3 m/s
valina [46]

Answer:

0.8 seconds

Explanation:

F=ma

Let x be the seconds the force is applied.

m = 20kg

F = 50 Newtons (kg*m/sec^2)

acceleration, a, is provided for x seconds to increase the speed from 1 m/s to  3 m/s, an increase of 2m/s

Let's calculate the acceleration of the cart:

F=ma

(50 kg*m/s^2) = (20kg)*a

a = 2.5 m/s^2

---

The acceleration is 2.5 m/s^2.  The cart increases speed by 2.5 m/s every second.  

We want the number of seconds it takes to add 2.0 m/sec to the speed:

(2.5 m/s^2)*x = 2.0 m/s

x = (2.0/2.5) sec

x = 0.8 seconds

7 0
2 years ago
The near point of an eye is 110 cm. A corrective lens is to be used to allow this eye to focus clearly on objects 26.0 cm in fro
irga5000 [103]

Answer:

The focal length of the lens is 34.047 cm

The power of the needed corrective lens is 2.937 diopter.

Explanation:

Distance of the object from the lens,u = 26 cm

Distance of the image from the lens ,v= -110 cm

(Image is forming on the other side of the lens)

Since ,lens of the human eye is converging lens,convex lens.

Using a lens formula:

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{26 cm}+\frac{1}{-110 cm}

f = 34.047 cm = 0.3404 m

Power of the lens = P

P=\frac{1}{f}=\frac{1}{0.34047 m}=2.937 diopter

6 0
3 years ago
The grant that considered the foundation of financial aid is the:
navik [9.2K]

Answer:

I think it is the Federal Pell Grant Program.

Explanation:

4 0
3 years ago
Read 2 more answers
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