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3 years ago

Which forces do not require contact between objects in order to exert a push or pull?

Physics

2 answers:

arsen [322]3 years ago

6 0

Answer:

Gravitational force and magnetic force. The other forces on the list require contact (tension requires contact with the rope and friction requires the object to be in contact with the surface)

Explanation:

Allisa [31]3 years ago

5 0

Answer:

gravitational force between the earth and an object and magnetic force between two magnets

Explanation:

gravitational force between the earth and an object and magnetic force between two magnets are forces that does not require contact between objects in order to exert a push or pull.

The earth and the object are never in contact with each other but there is a force of gravitation existing between them. There is force of attraction between the masses of the earth and the masses according to gravitational law are separated by a distance.

Force between two magnets may or maynot exert push or pull. They only attract when the poles of the magnets are unlike poles (different). There is force of repulsion between them if like poles of the magnet come together according to the law of magnetism. Magnets in magnetic field will always experience force even if at a distance from each other in the field hence the reason for the selected options

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Which statement describes the relationship between voltage and current

DaniilM [7]

Answer:

Ohm's Law. The relationship between voltage, current, and resistance is described by Ohm's law. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r.

Explanation:

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4 years ago

A skier with a mass of 63 kg starts from rest and skis down an icy (frictionless) slope that has a length of 50 m at an angleof

victus00 [196]

Answer:

Explanation:

When the skier reaches the bottom of the slope , height lost by it

h = 50 sin32 m

= 26.5 m

potential energy lost

= mgh

Gain of kinetic energy

= 1/2 mv²

mgh = 1/2 mv²

v = √ 2gh

= √ (2x9.8 x 26.5)

= 22.8 m /s

b )

Let μ be the coefficient of kinetic friction required.

friction force acting

= μmg

work done by friction in displacement of d (40 m ) on horizontal surface

- μmg x d

This negative work will be equal to positive kinetic energy of the skier on horizontal surface .

= μmg x d = (1/2) m v²

μ = v² / (2 gd)

= 519.4 / (2 x 9.8 x 140 )

= .19

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4 years ago

Why is weight considered a force

saul85 [17]

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4 years ago

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5 0

3 years ago

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

3 0

3 years ago