A 55.2 kg softball player moving 3.11 m/s slides across dirt with uk=0.310. How far does she slide before coming to a stop

If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a resul

Kobotan [32]

-- If the system is 'closed', then nothing ... including energy ... can get in or out, and the total energy inside has to be constant.

If half of the energy in the system starts out as potential energy and the rest starts out as kinetic, and then the potential energy increases, there's only one place the increase could have come from ... it could only have been converted from kinetic energy. So the kinetic energy in the system must decrease.

In fact, this isn't even a "result". The kinetic energy has to decrease before the potential energy can increase, because that's where the increase has to come from.

If the system is 'open', then energy can come in and go out. If the potential energy inside suddenly increases, we don't know where it came from, so we can't say anything about what happens to the system.

7 0

3 years ago

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

mars1129 [50]

(M G H)=(0.5 x 9.8 x 10) = 49 joules.

4 0

3 years ago

North Dakota Electric Company estimates its demand trend line (in millions of kilowatt hours) to be: D = 75.0 + 0.45Q, whe

Alborosie

Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line is

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

We need to calculate the demand forecast for winter

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times101)\times0.80$

$D=96.36\ millions\ KWH$

We need to calculate the demand forecast for spring

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times102)\times1.20$

$D=145.08\ millions\ KWH$

We need to calculate the demand forecast for summer

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times103)\times1.40$

$D=169.89\ millions KWH$

We need to calculate the demand forecast for fall

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times104)\times0.60$

$D=73.08\ millions KWH$

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

3 0

3 years ago

1. The process of producing energy by utilizing heat trapped inside the earth's

babymother [125]

Answer:

B

Explanation:

The process is called Geo-Thermal energy because it's an inexhaustible source of energy

6 0

3 years ago

Read 2 more answers

CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY!

BARSIC [14]

<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />

According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is the acceleration of gravity.

Now, in this case we are on Earth, so the gravity value is $9.8\frac{m}{s^{2}}$

Note the objects experience the acceleration of gravity regardless of their mass.

Nevertheless, on Earth we have air, hence air resistance, so the afirmation "Free fall is a situation in which the only force acting upon an object is gravity" is not completely true on Earth, unless the following condition is fulfiled:

If the air resistance is too small that we can approximate it to zero in the calculations, then in free fall the objects will accelerate downwards at $9.8\frac{m}{s^{2}}$ and hit the ground at approximately the same time.

5 0

4 years ago

A 55.2 kg softball player moving 3.11 m/s slides across dirt with uk=0.310. How far does she slide before coming to a stop​

A 55.2 kg softball player moving 3.11 m/s slides across dirt with uk=0.310. How far does she slide before coming to a stop