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Juliette [100K]
2 years ago
5

A ball is attached to a vertical spring. The ball is initially supported at a height y so that the spring is neither stretched n

or compressed. The ball is then released from rest and it falls to a height y - h before moving upward. Consider the following quantities: translational kinetic energy, gravitational potential energy, elastic potential energy. When the ball was at a height y - (h/2), which of the listed quantities has (have) values other than zero joules
Physics
1 answer:
scoundrel [369]2 years ago
3 0

Answer:

All the three quantities will have non zero joules.

Explanation:

At the initial position of rest the system will have only gravitational potential energy while the other 2 quantities will be zero.

when the system reaches a height (y-h) only kinetic energy will be zero while the other 2 quantities will be non zero

At the position of (y-h/2) all the 3 quantities will be non zero.

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Technician A says that if the opposite DTC can be set, the problem is the component itself. Technician B says that if the opposi
Fiesta28 [93]

Answer:Both are correct

Explanation:

The opposite DTC also comprises of the wiring or ground. If the opposite DTC can be set it is the components that is faulty and if otherwise it is still the components that is faulty

5 0
3 years ago
The Position Of A Particle Moving In The Xy Plane Is Given By R= [2.0*cos(3.0t)i+ 2.0*sin(3.0t)j] Where R Is In Meters And T Is
ExtremeBDS [4]

Answer:

Explanation:

We can prove that this is a circular motion if we demonstrate that the norm of the vector is independent of time. Hence we have

R(t)=2.0cos(3t)\hat{i}+2sin(3t)\hat j\\\\|R(t)|=\sqrt{4cos^2(3t)+4sin^2(3t)}\\\\|R(t)|=\sqrt{4(cos^2(3t)+sin^2(3t))}\\\\

but cos^2(3t)+sin^2(3t)=1. Thus we obtain

|R(t)|=\sqrt{4}=2

The norm is independent of t, thus, the particle describes a circular motion

Hope this helps!!

3 0
2 years ago
Room's become bright at the day time although there is no bright sunlight in the rooms why
Aneli [31]

Answer:

The room become bright in the day by a process call scattering of light or Rayleigh scattering.

Explanation:

It is called Rayleigh scattering or scattering of light because Rayleigh scattering is the scattering if light or electromagnetic radiation by smaller particles which have radius of less than 110 nanometer in a medium and the wavelengths of the electromagnetic. Wave or light remain unchanged. The scattering of light occur in the day time in the room and this bring of brighten up of the room.

5 0
2 years ago
Need help in science <br><br> Please help me pass my exam hw
iren2701 [21]
Well I'm in eight and I do high school-level homework/schoolwork. And yes, your question has been answered but if you need help next time I'm free to help!
4 0
2 years ago
A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i
IrinaVladis [17]
A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
W=Fd = (46.3 N)(4.25 m)=196.8 J

b) The work done by the frictional force against the motion of the block is equal to:
W_f =  -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=
=-105.1 J
Part of these 105.1 Joules of work becomes increase of thermal energy of the block (\Delta E_B), and part of it becomes increase of thermal energy of the floor (\Delta E_F). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J

c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
W_{net}=W-W_f=196.8 J-105.1 J=91.7 J
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
W_{net} = \Delta K
So, we have \Delta K=+91.7 J


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3 years ago
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