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3 years ago

11

What is the momentum of a 0.2kg ball with a speed of 5 m/s

1 answer:

Eddi Din [679]3 years ago

3 0

Momentum

= mass x velocity

= 0.2 x 5

= 1 kg m/s

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A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an

agasfer [191]

Answer:

$t=\sqrt{h/g}$

Explanation:

We use the kinematics equation to solve this question:

$y(t)=y_{o}+v_{o}t+1/2*a*t^{2}$

$v_{o}=0$ because the ball is dropped

$a=-g$ the acceleration is the gravity, negative because it points downwards

$y_{o}=h$ initial height

$y(t)=h/2$ final height

So:

$h/2=h-1/2*g*t^{2}$

$t=\sqrt{h/g}$

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3 years ago

Which of the following is a form of potential energy? O A. Sound energy O B. Elastic energy O C. Light energy O O O D. Kinetic e

olga55 [171]

Answer:

O C. Light energy

Explanation:

it conducts energy in it and is an energy itself.

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2 years ago

What is the tangent function used to find and what are its units

Verizon [17]

The functions of angles are used to find unknown lengths or angles that can't be measured, in terms of known quantities. The trig functions of angles are ratios of lengths, so they're bare naked numbers without units.

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3 years ago

A 0.54 kg bullfrog is sitting at rest on a level log. How large is the normal force of the log on the bullfrog?

RUDIKE [14]

Answer:

<h2>5.3N</h2>

Explanation:

Step one:

given data

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Step two:

Required

The force F in Newton

From newton first law

F=mg

The acceleration due to gravity acting on the frog is 9.81m/s^2

hence

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3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

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3 years ago