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3 years ago

Distinguish between contact force and non-contact forces give three examples of each

Physics

1 answer:

Artist 52 [7]3 years ago

6 0

Answer:

1. the force which can be felt or act only when two objects are in contact is known as contact force.

for example: frictional force, muscular force,

tension, air resistance .

2. the force which can be felt or act even when two objects are in contact or not is known as non-contact force.

for example: magnetic force, gravitational force, electrostatic force.

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arsen [322]

The weight is not coming from the center of the mass because the force that act on it is not is equal is side.(2) section B donot have weight because the ruler bend down and section be raise up so no weight.

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3 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

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3 years ago

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3 years ago

Read 2 more answers

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

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3 years ago

Technician A says a short to voltage may or may not blow a fuse and may or may not affect more than one circuit. Technician B sa

Tomtit [17]

Answer:

C. Both A and B

Explanation:

Fuses are rated by the amperage they can carry before heat melts the element. The fuse is ideal for protection against short circuits. Short circuits produce enough amperage to vaporize a fuse element and break connection in one cycle of a 60-cycle system.

Specifically, the voltage rating determines the ability of the fuse to suppress the internal arcing that occurs after a fuse link melts and an arc is produced.

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3 years ago