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gavmur [86]
3 years ago
10

A certain spring stores 10.0J of potential energy when it isstretched by 2.00cm from its equilibrium position.How much potential

energy would the spring store if it werestretched an additional 2.00cm?How much potential energy would it store if it were compressedby 2.00cm from its equilibrium position?How far from the equilibrium position would you have tostretch the string to store 20.0J of potential energy?What is the force constant of this spring?
Physics
1 answer:
qaws [65]3 years ago
3 0

Answer:

Answered

Explanation:

x= 0.02 m

E_p= 10.0 J

E_p= 0.5kx^2

10= 0.5k(0.02)^2

solving we get

K= 50.0 N/m

Now

E'_p= 0.5kx'^2

E'_p= 0.5×50×(0.04)^2

E'_p=40 J

b) potential energy is a scalar quantity and it only depends magnitude and not direction so it will remain same in compression and expansion both

c) 20 J = 0.5×50,000×x^2

solving

x= 0.028 m

d) k is 50.0 N/m  from above calculation

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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
Guys No one's answering my question so sad! Once again I'm asking the same question –Here
nexus9112 [7]

For the front glass of the car to get wet, V_c \geq 10 \ m/s.

The given parameters:

  • <em>Speed of the car, = Vc</em>
  • <em>Speed of the rain, = 10 m/s</em>

The relative velocity of the car with respect to the falling rain is calculated as;

V_{C/R} = V_C- V_R

  • If the speed of the car equals the speed of the rain, the rain will fall behind the car.
  • If the speed of the rain is greater than speed of the car, the rain will fall far in front of the car.
  • If the speed of the car is greater than speed of the rain, the rain will fall on the car.

Thus, for the front glass of the car to get wet, V_c \geq 10 \ m/s.

Learn more about relative velocity here: brainly.com/question/17228388

8 0
2 years ago
You are leaving for a party at your cousin’s house in a city that is 150 km away. You will travel at an average rate of 50 km/hr
choli [55]
3 hours, because for every 50 km equals one hour 150 divided into 50 equals 3
5 0
3 years ago
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A sample of aluminum has a mass of 4.86 grams and a volume of 1.8 cubic centimeters. What is the density of aluminum rounded to
TEA [102]

Answer:

3g/cm³

Explanation:

2.7

The seven adds 1 to the 2 making it 3.0

6 0
3 years ago
Part CPart complete Determine the maximum force P that can be applied without causing the two 43-kg crates to move. The coeffici
Wewaii [24]

Answer:

the maximum force will be equal to 134.84 N  

Explanation:

We have given mass m = 43 kg

Coefficient of static friction \mu _s=0.32

Acceleration due to gravity g=9.8m.sec^2

We have to find the maximum force which , when applied there is no movement of crates

This maximum force will be equal to frictional force

Frictional force is given by f=\mu mg=0.32\times 43\times 9.8=134.84N

So the maximum force will be equal to 134.84 N  

3 0
3 years ago
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