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3 years ago

Which situation is contrary to Newton’s first law of motion?

Physics

1 answer:

Bond [772]3 years ago

4 0

Answer:

An object at rest stays at rest as long as unbalanced forces act on it.

Explanation:

Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.

In physics, Sir Isaac Newton's First Law of Motion is known as Law of Inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.

The inertia of a physical object such as a truck is greatly dependent or influenced by its mass; the higher the quantity of matter in a truck, the greater will be its tendency to continuously remain at rest.

Hence, the situation which is contrary to Newton’s first law of motion is that, an object at rest stays at rest as long as unbalanced forces act on it.

According to Newton’s first law of motion, an object at rest stays at rest as long as unbalanced forces do not act on it.

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A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium

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Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

w² = $\frac{k}{m}$

to find the constant (k) of the spring, we use Hooke's law with the initial data

F = - kx

where the force is the weight of the body that is hanging

F = W = m g

we substitute

m g = - k x

k = $- \frac{m g}{x}$

we calculate

k = $- \frac{9.8 m}{- 2.6 \ 10^{-2}}$

k = 3.769 10² m

we substitute in the first equation

w² = $\frac{ 3.769 \ 10^2 \ m }{m}$

w = 19.415 rad / s

angular velocity and frequency are related

w = 2πf

f = $\frac{w}{2\pi }$

f = 19.415 / 2pi

f = 3.09 Hz

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3 years ago

5. What structure did Friedrich Kekule discover that allowed carbon atoms can bond

musickatia [10]

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3 years ago

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of

Gekata [30.6K]

Answer:

a) $P=2450\ Pa$

b) $\delta h=23.162\ cm$

Explanation:

Given:

height of water in one arm of the u-tube, $h_w=25\ cm=0.25\ m$

Gauge pressure at the water-mercury interface,:

$P=\rho_w.g.h_w$

we've the density of the water $=1000\ kg.m^{-3}$

$P=1000\times 9.8\times 0.25$

$P=2450\ Pa$

Now the same pressure is balanced by the mercury column in the other arm of the tube:

$\rho_w.g.h_w=\rho_m.g.h_m$

$1000\times 9.8\times 0.25=13600\times 9.8\times h_m$

$h_m=0.01838\ m=1.838\ cm$

<u>Now the difference in the column is :</u>

$\delta h=h_w-h_m$

$\delta h=25-1.838$

$\delta h=23.162\ cm$

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