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3 years ago

Force that opposes motion between two surfaces

Physics

1 answer:

RSB [31]3 years ago

4 0

Frictional force

Explanation:

Frictional force is a force that opposes the motion between two surfaces that are in contact with one another.

Frictional force opposes motion and it is directed opposite the direction of motion.

There is also frictional force between liquids too.

Enough force must be supplied in order to overcome this frictional force so as to make a body move.
In humans, motion wouldn't be possible without the frictional force.

learn more:

Friction brainly.com/question/7174363

#learnwithBrainly

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Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long

nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

$R = \dfrac{\rho L }{A}$

where;

A = πr²

$R = \dfrac{\rho L }{\pi r ^2}$

For the shorter cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{ 4 \rho L }{\pi \ D ^2}$

For the longer cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{32\rho L }{\pi \ (4 D) ^2}$

$R = \dfrac{2\rho L }{\pi \ (D) ^2}$

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

$\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}$

$\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}$

$\dfrac{R_s}{R_L} =2$

${R_s}=2{R_L}$

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0

3 years ago

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$