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Shtirlitz [24]
3 years ago
9

Write a paragraph about pull.

Physics
2 answers:
Zanzabum3 years ago
6 0

Answer:A force can be described as a push or a pull. Pushes and pulls can be seen to act on objects when they begin to move, speed up, slow down or change direction.

The image gallery on this page shows children at play using push and pull actions to move objects. Children use push and pull all the time while at play. While playing, students learn to manipulate objects and materials and make observations about their actions.

Teachers may use this teaching resource over a number of lessons to explore and develop their students' understanding that a push or a pull affects how an object moves or changes shape.

Explanation:

vovikov84 [41]3 years ago
3 0
In physics, we speak of forces applied to objects. This is all part of Newtonian mechanics. The key is that an object at rest will remain at rest unless a force acts on it, and an object in motion will continue in motion at the same speed (velocity) and in the same direction unless a force acts on it.

Forces can be categorized in various ways. A pull would be a force towards that which is pulling. A push would be a force away from that which is pulling.

Forces can also be categorized by whether contact is needed for the force to act on the object. An example of a pulling force with contact would be where I attach a rope to something and pull it towards me. Pulling can also be called attraction. An example of a pushing force with contact would be a billiard ball being hit by the cue ball and moving away from the cue ball. Pushing can be called repulsion.

There are also forces that act with no contact. The force of gravity is an attractive force only. Things fall towards the Earth when we let go of them due to gravity, and the earth stays in orbit around the sun instead of flying away in a straight line due to gravity. There is no gravity of repulsion, where objects are pushed apart due to their mass. Or at least, we haven't found one yet.

Electromagnetic force can attract (pull) or repel (push). Two magnet will attract from the north pole of one to the south pole of the other, but repel if the alignment is north-north or south-south.

Many forces can act at once, and forces come at odd angles, too. A cue ball will repel the ball it hits. But only if it hits straight on (with the line of travel of the cue ball the same as the line from the center of the cue ball to the center of the ball being hit) will the ball being struck move straight away from the cue ball. If the collision is at an angle, the push will occur, but it will be at an angle, and only some of the force will transfer.

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Answer:

HCP

Extra -> BCC the most, FCC between

Explanation:

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A cannonball is launched from the ground with 48279 Joules of kinetic energy. How much gravitational potential energy will the c
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When the cannonball is shot upwards, kinetic energy converts to gravitational potential energy (GPE).

As the ball rises, speed decreases and height increases.

So when the cannonball has reached its maximum height, all of the Kinetic energy has transferred into gravitational potential energy.

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So the GPE is 48279 Joules at it's maximum height.

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3 years ago
Read 2 more answers
A mass m at the end of a spring vibrates with a frequency
Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

F = -kx

And Newton's Second Law also states that

F = ma = m\frac{d^2x}{dt^2}

Combining two equations yields

a = -\frac{k}{m}x

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

\omega = \sqrt{\frac{k}{m}}

And given that ω = 2πf

the relation between frequency and mass becomes

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Let's apply this to the variables in the question.

0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg

6 0
3 years ago
What is the electric potential difference of 5cm from a point charge of -2.0µc
vodka [1.7K]

Answer:

V = 360 kV

Explanation:

Given that,

Charge, q = -2µC

We need to find the electric potential difference of 5cm from a point charge.

We know that the formula for the electric potential is given by :

V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 2\times 10^{-6}}{0.05}\\\\V=360000\ V

or

V=360\ kV

So, the electric potential difference is 360 kV.

7 0
3 years ago
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