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3 years ago

A load of 36 N attached to a spring hanging

Physics

1 answer:

givi [52]3 years ago

3 0

Answer:

The required force is 55.38 N.

Explanation:

First we have to find the spring constant K;

k = $\frac{mg}{d}$ = $\frac{36}{0.065}$ = 553.846 N/m

As,

F= -K $x$

or, F= - 553.846 × 0.10

or, F= -55.3846 N

Hence,

$F_{needed}$ = - $F_{spring}$

∴ $F_{needed}$ = 55.38 N

The required force to stretch it by this amount is 55.38 N.

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A student pushes a 40-N block across the floor for a distance of 10 m. How much work was applied to move the block? A. 4 J B. 40

damaskus [11]

Now for this problem, what is given is a 40 Newtons which would represent the force to be applied to the object, and a distance of 10 meters after the application of the said force. When these two combine, work is done. The unit for work is Joules and this is what we are looking for. The formula to get Joules or for work would be the force applied to the object multiplied by the distance that it travelled after the application of the force. It looks like this

work = force x distance
Joules = Newtons x meter

so let us substitute the variables to their corresponding places

Joules = 40 N x 10 m
Joules = 400 J

So the answer to this question would be C. 400 J

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maw [93]

Density is mass over volume. Density is also a charecteristic property wich defines a substance and two different substance cant have the same charasteristic property.

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4 years ago

What happened if there is no frictional force?

Komok [63]

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3 years ago

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4 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$