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2 years ago

6

Hi i do not have a question i just want to show u my cute doggo :)) ik everyone on here is probably stressed so hope this makes

ur day better
btw this is not my dog but if u see the first one i did that is my dog heheh

2 answers:

Arisa [49]2 years ago

5 0

Answer:

wooowwwwww

Explanation:

it's so so cuteeeeeeee

I love dogs tho:))

alekssr [168]2 years ago

3 0

Answer:

cute dog can I get brainly

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2 years ago

What do you understand by upthrust Force

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Answer:

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Explanation:

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2 years ago

This whole worksheet. I am not good at this and I am completely lost

Ket [755]

1.) potential energy
2.)potential and kinetic
3.)The roller coaster car has the most kinetic energy at point X i know this because the car is moving and kinetic energy has the power to move or change things therefore point X is when the roller coaster car has the most energy.
4.)potential energy
5.)kinetic energy
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2 years ago

Water flows over a section of Niagara Falls at the rate of 1.4 × 106 kg/s and falls 49.8 m. How much power is generated by the f

Ad libitum [116K]

Answer:

Power= 6.84×10⁸ W

Explanation:

Given Data

Niagara falls at rate of=1.4×10⁶ kg/s

falls=49.8 m

To find

Power Generated

Solution

Regarding this problem

GPE (gravitational potential energy) declines each second is given from that you will find much the kinetic energy of the falling water is increasing each second.

So power can be found by follow

Power= dE/dt = d/dt (mgh)

Power= gh dm/dt

Power= 1.4×10⁶ kg/s × 9.81 m/s² × 49.8 m

Power= 6.84×10⁸ W

7 0

2 years ago

Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above

poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$W_{dis}$ - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

$f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}$

If $m = 1000\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 40\,m$ , $y_{2} = 25\,m$ and $\Delta s = 400\,m$ , then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0

2 years ago