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Citrus2011 [14]
2 years ago
15

A object is placed between the focal point and the lens of a concave lens. Where will the image be formed?

Physics
1 answer:
Vesnalui [34]2 years ago
6 0

Answer:

When object is placed between convex lens and focal point, image formed is virtual and magnified which is shown in the diagram. Was this answer helpful?

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how does electromagnetism effect the function of a generator (using magnetism to produce electrical energy)
Sati [7]

Explanation:

We know that a changing magnetic field induces a current in a conductor. For that reason a generator basically consist an element that produces a magnetic field that changes over time and a conductor where the current will be induced.

This element that produces a magnetic field can be one of the following:

- A permanent magnet: Which is basically like a regular magnet. The magnetic field that a permanent magnet produces does not change over time, we need a motor or any other external force to move the axis of the generator and cause the magnetic field to change.

- An electro-magnet. Which is basically a DC current flowing through a conductor. Basically, when current flows through a conductor it behaves exactly like a magnet. So what we commonly do, is to connect a conductor to a DC battery, and it will create a magnetic field.

Like we are using a DC battery to create a magnetic field, then the magnetic field won't change over time either. So we still need an external force to move the axis of the generator to produce AC electricity.

4 0
3 years ago
Which type of electromagnetic waves has highest frequency​
vesna_86 [32]

Answer:

Gamma rays have the highest energies.

Explanation:

HOPE IT WILL HELP ^_^

6 0
3 years ago
Read 2 more answers
Samuel adds a teaspoon of salt to a glass of water. He notices that the salt disappears. Samuel takes a dip to discover that the
Tpy6a [65]
It is a chemical change because the salt is dissolving.
6 0
2 years ago
a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi
zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

                     v = u + at

                     a = (v - u) / t

Substituting the given values in the above equation

                    a = (10 - 0) / 3

                       = 3.33 m/s²

Using the second equations of motion

                      s = ut + 1/2 at²

                         = 0 + 0.5 x 3.33 x 3²

                         = 15 m

The force exerted by the car

                         F = m x a

                            = 1500 Kg x 3.33 m/s²

                            = 5000 N

The work done by the car,

                          W = F x S

                               = 5000 N x 15 m

                               = 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

                            P = W / t

                                = 75,000 J / 3 s

                                 = 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0
3 years ago
A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri
exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
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