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WINSTONCH [101]
2 years ago
15

Applying the Law of Conservation of Energy. If a car was released down the track from a height what happens to the potential ene

rgy and Kinetic energy as the car goes down the ramp?
Physics
1 answer:
erastova [34]2 years ago
4 0

Answer:

According to the law of conservation of energy, energy cannot be created or destroyed,  although it can be changed from one form to another.    KE + PE = constant. A simple example involves a stationary car at the top of a hill.  As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases.  On the way back up the hill, the car converts kinetic energy to potential energy.  In the absence of friction, the car should end up at the same height as it started.

This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.

One can also imagine the energy transformation in a pendulum.  When the ball is at the top of its swing, all of the pendulum’s energy is potential energy.   When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy.   The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms

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Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred
madam [21]

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K   ,V₁ = 3 m³  ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas  P V = m R T

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

P_2=\dfrac{T_2}{T_1}\times P_1

P_2=\dfrac{600}{300}\times 2

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

m=\dfrac{PV}{RT}

m=\dfrac{200\times 3}{0.287\times 300}\ kg

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0  Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}

S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}

S₂ - S₁ = 3.42 KJ/K

 

5 0
3 years ago
two horses pull against a rope with forces of 100 newtons in opposite directions. this is an example of
statuscvo [17]
It is an example of balanced force.


hope this helps. good luck 
4 0
3 years ago
Read 2 more answers
A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti
LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision = 7.668\ ms^-1

Their common speed after the collision = 2.56\ ms^-1

Height achieved by the package of mass m when it rebounds = 0.33\ m

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

where K is Kinetic energy and U is Potential energy.

K= \frac{mv^2}{2} and U= mgh

Considering the fact  K_{initial} = 0\ and U_{final} =0 we will plug out he values of the given terms.

So V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1

Keypoints:

  • Sum of energies and momentum are conserved in all collisions.
  • Sum of KE and PE is also known as Mechanical energy.
  • Only KE is conserved for elastic collision.
  • for elastic collison we have e=1 that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where V_1{i} =7.668\ ms^-1.

Plugging the values we have

m\times 7.668 = (3m)\times V_{f}

Cancelling m from both sides and dividing 3 on both sides.

V_f = 2.56\ ms^-1

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of V_{f} for m mass.

As here V_{f}=-2.56\ ms^-1 we can use that if both are moving in right ward with 2.56 then there is a  -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object 1.

So

V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1

Now using law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh

\frac{v(f1)^2}{2g} = h

h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed =7.668\ ms^-1 for part b we have their common speed =2.56\ ms^-1 and for part c we have the rebound height =0.33\ m.

3 0
3 years ago
What is the equivalent resistance if you connect three 10.0 Ω resistors in series?
Sonja [21]

Req = 30.0Ω.

When two or more resistors are in series, the intensity of current that passes through each of them is the same. Therefore, if you notice, you can observe that the three previous series resistors are equivalent to a single resistance whose value is the sum of each one.

Req = R1 + R2 + R3 = 10.0Ω + 10.0Ω + 10.0Ω = 30.0Ω

4 0
3 years ago
Read 2 more answers
The format specifier ________ is a placeholder for an int value. %d %n %int %s
Grace [21]

%d is a format specifier that is a placeholder for an int value. It tells the compiler that we want to print an integer value that is present in variable a. In this way there are several format specifiers in c.

8 0
3 years ago
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