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o-na [289]
2 years ago
9

A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq

uilibrium. what is the work done on the block by the spring.
A. W=kx^2
B.W=-kx^2
C.W=0
D. None of these
Physics
1 answer:
Snowcat [4.5K]2 years ago
6 0

Answer:d

Explanation:

Spring is compressed to a distance of x from its equilibrium position

Work done by block on the spring is equal to change in elastic potential energy

i.e. Work done by block W=\frac{1}{2}kx^2

therefore spring will also done an equal opposite amount of work on the block in the absence of external force

Thus work done by spring on the block W=-\frac{1}{2}kx^2

Thus option d is correct

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A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
dlinn [17]

Answer:

a = 4.72 m/s²  

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a                        

f is the friction force

m g sinθ - μ N = ma                        

m g sinθ - μ m g cos θ = ma            

a = g sinθ - μ g cos θ                    

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²                                      

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²  

3 0
2 years ago
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False
Studentka2010 [4]
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is? 
B.6

3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
4 0
2 years ago
Read 2 more answers
Discuss the production, transmission, and usage of electricity in the context of conservation of energy. When electricity is “us
mr Goodwill [35]

There are huge losses in the transmission, production and usage of electricity and the reduction of these losses in order to save electricity is called as conservation of energy.

As per the statistics, there is loss of nearly 4% while the transmission of electricity. Like wise during production also, lot of electricity get wasted due to the inefficient material used. None of the production material nor the equipment used have 100% efficiency and thus there is always a possibility of energy wastage.

When it is said that the energy is wasted , it simply means that the energy production which should have been 100% as per calculation is not completely derived from the source due to the inefficient conversion process. For example, a turbine while rotating must convert 100 % of the water energy or water falling on it into electrical energy but the turbine is not able to do so as some of the water is lost or its energy is lost before conversion while going through the mechanical process.

3 0
2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus
dexar [7]

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

6 0
2 years ago
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