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3 years ago

Define Momentum in detail.

Physics

2 answers:

Harrizon [31]3 years ago

8 0

Explanation:

Momentum Is defined as the product of of mass and its velocity

Momentum (M) =mass *velocity

SI unit of momentum is kgm/s

The rate of change in momentum

=change in momentum / time

=(mv-mu)/t

blsea [12.9K]3 years ago

7 0

Answer:

Momentum is simply the product of mass and velocity. It is represented mathematically as

Momentum = Mass x Velocity.

The S. I unit of momentum is kilograms meter per second i.e Kg.m/s.

Momentum is a vector quantity i.e it has both magnitude and direction.

Momentum of an object is greatly affected my the mass and velocity of the object as we can clear from the equation.

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Physical Science is the study of

vazorg [7]

Answer:

Physical science is the study of the inorganic world. That is, it does not study living things. (Those are studied in biological, or life, science.) The four main branches of physical science are astronomy, physics, chemistry, and the Earth sciences, which include meteorology and geology.

Explanation:

6 0

3 years ago

Read 2 more answers

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

4 0

3 years ago

Pls help meh Kdjdjeidjndiejdididjdjjdidjdjdiejd

Nadya [2.5K]

Answer:

i think b

Explanation:

i'm not sure

6 0

2 years ago

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Why can even a small injury to the cornea have a major effect on vision?

Rudik [331]

Your eyes are the most delicate thing on your body. Little things can make you go blind

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3 years ago

Part 1: Use complete sentences to explain why solar winds occur. Part 2: Give two examples in which solar winds impact Earth.

Troyanec [42]

Part 1

When the solar atmosphere accumulates a lot of magnetic energy to a point that cannot accumulate more, all that magnetic energy is suddenly released, and with it, a lot of radiation. So much, that in fact it covers all of the electromagnetic spectrum; from radio waves to gamma rays. That burst of radiation is called a solar flare. In a single solar flare the amount of radiation released is millions of times greater than all the nuclear bombs in the face if the earth exploding together. Lucky for us, most of the high-energy radiation dissipates before reaching the Earth, and the radiation that do reach us, is deflected by the Earth’s magnetic field.

Part 2

1. Not all the radiation of solar flares that reach the Earth is deflected by its magnetic field; some of them reach us and charges the upper atmosphere with ionized particles. Those particles react with the gases in the atmosphere and produce a light; that light is what we call Auroras borealis or southern nights; One the most beautiful natural spectacles in earth, who thought Auroras begin their lives as deadly solar flares.

2. Solar flares contain a lot of high-energy radiation that is extremely dangerous for our electronic devices; when they reach the Earth, they can damage sensible electronics like satellites. A very powerful solar flare could even damage all the electronic devices on the surface of the Earth.

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2 years ago