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4 years ago

What is the equation for gravity and what does each part mean?

1 answer:

6 0

This equation description the force between any two object in the universe in the equation F is the force of the gravity Newton lawe

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5)A 0.50 kg hockey puck is at rest on ice when you hit it with a hockey stick, applying a force of 100 N for

mojhsa [17]

Answer:

F t = m Δv impulse delivered = change in momentum

Δv = 100 * .1 / .5 = 20 m/s original speed of puck

KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J initial KE of puck

E = μ m g d energy lost by puck

Ff = μ m g = m a deceleration of puck due to friction

a = μ g = 9.8 * .2 = 1.96 m/s^2

v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s speed of puck on striking box

m v2 = M V conservation of momentum when puck strikes box

V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s speed of box after collision

KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J KE of box after collision

KE = μ M g d energy lost by box in sliding distance d

d = 23.3 / (.3 * .8 * 9.8) = 9.91 m distance box slides

7 0

2 years ago

J. J. Thomson s experiments provided evidence that an atom

gulaghasi [49]

J.J. Thomson's experiments provided evidence that an atom B.contains negatively charged particles.

<span>♡♡Hope I helped!!! :)♡♡
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4 years ago

When the rock or soil of Earth’s surface is moved to another location by water, ice, or wind

levacccp [35]

The answer is deposition/A. Please mark brainliest.

8 0

3 years ago

Read 2 more answers

29:45

Yakvenalex [24]

Answer:

it opens the circuit so that electric charges do not flow to the timier

Explanation: a closed circut will allow electrisity to flow through.

6 0

3 years ago

Read 2 more answers

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

3 years ago