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Zanzabum
3 years ago
13

What are two ways to increase frictional force?

Physics
1 answer:
Mekhanik [1.2K]3 years ago
7 0
Ways to increase friction - increase the area of contact - increase the roughness of the contact materials - dry up or take away any lubricant - increase the pressure on the contact
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Explain why wedges and screws are actually types of inclined planes.
slava [35]
<span>The screw is really just an inclined plane covered around with a tiny pole. The wedge is definitely an inclined plane, since it starts with a point, then rises getting thicker, as an inclined plane. </span>
5 0
4 years ago
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Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area
yaroslaw [1]

Answer:

5.63\cdot 10^{-6} C

Explanation:

The capacitor of a parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

A is the area of each plate

d is the separation between the plates

\epsilon_0 is the vacuum permittivity

The energy stored in a capacitor instead is given by

U=\frac{1}{2}\frac{Q^2}{C}

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}

And re-arranging it

Q=\sqrt{\frac{2U\epsilon_0 A}{d}}

Now if we substitute

d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J

We find the charge stored on the capacitor:

Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C

7 0
3 years ago
Which of these is NOT an example of a reference direction?
Vikentia [17]

Answer:

A

Explanation:

8 0
4 years ago
What is air pressure?
Art [367]
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8 0
3 years ago
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A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.
fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe,  the velocity is given by

v=\frac {2Lf}{n}

Making L the subject then

L=\frac {nV}{2f}

Where f is the frequency,  L is the length,  n is harmonic number,  v is velocity

Substituting 1 for n,  320 Hz for f and 331 m/s for v then

L=\frac {1*331}{2*320}=0.5171875\approx 0.52m

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, f2=2\times \frac {v}{2L} and f3=3\times \frac {v}{2L}

f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz

(c)

When v=367 m/s then

f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz

5 0
3 years ago
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