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3 years ago

Chromatic aberration occurs when:___.

Physics

1 answer:

worty [1.4K]3 years ago

4 0

Answer:

b. a lens does not focus all colors of light to the same place.

Explanation:

Chromatic aberration is a defect of a lens. In this defect, the lens is unable to focus the different wavelengths of the light on a single focal point. It is also known as chromatic distortion and color fringing. It is caused by the dispersion of light while passing through a lens. As a result, the image might become blurred and different colors are observed around its edges. It can be corrected by the use of a combination of converging and diverging lenses.

Hence, the correct option will be:

b. a lens does not focus all colors of light to the same place.

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if a boat was moving at a velocity of 30 m/s to the south and it sailed for 30 minutes, what was the displacement of the boat?

kifflom [539]

If you move south at 30 m/s for 30 minutes, your displacement is

D = (30 mtr/s) x (30 min) x (60 sec/min)

D = (30 x 30 x 60) (mtr-min-sec/sec-min)

D = 54,000 meters south

D = 54 km south

8 0

3 years ago

F Penny's piston is wider than Pauly's, who is pushing with greater force?

Anastasy [175]

A Penny is exerting the greater force.
B Pauly is exerting the greater force.
C They are both using equal force.
Think its A or C

8 0

3 years ago

Read 2 more answers

8. Two positive charges (+8.0 mC and +2.0 mC) are separated by 300 m. A third charge is placed at distance r from the +8.0 mC ch

muminat

Answer:

0.20 km

Explanation:

The computation of the distance r is shown below:

This can be calculated by using the following formula

As we know that

$\frac{k_q_1}{r^2} = \frac{k_q_1}{(300 - r)^2}$

where,

q_1 is 8 mC

q_2 is 2 mC

Now placing these values

Now

$\frac{k (8 \times 10^{-3c})}{r^2} = \frac{k (2 \times 10^{-3c})}{(300 - r)^2}$

$\frac{m}{r^2} = \frac{1}{(300 - r)^2}$

$\frac{2}{r} = \frac{1}{300 - r}$

r = 600 - 2r

r = 200 m

r = 0.20 km

We simply applied the above formula to determine the distance of r

Hence, the distance r is 0.20 km

4 0

3 years ago

20 POINTS.

ololo11 [35]

Answer:

i think its student b (d)

Explanation:

it might be student b cuz every action creates an equal and opposite reaction

7 0

3 years ago

A worker pushes a 50 kg crate a distance of 7.5 m across a level floor. He

Taya2010 [7]

a) 73.5 N

b) 551.3 J

c) -551.3 J

d) 0 J

e) 0 J

f) 0 J

g) 0 J

Explanation:

There are two forces acting on the crate:

- The push of the worker, F, in the forward direction

- The frictional force, $F_f=\mu mg$ , in the backward direction, where

$\mu=0.15$ is the coefficient of friction

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:

$F-F_f=ma$

However, the crate here is moving with constant velocity, so its acceleration is zero:

$a=0$

So the previous equation becomes:

$F-F_f=0$

And we can find the magnitude of the applied force:

$F=F_f=\mu mg=(0.15)(50)(9.8)=73.5 N$

The work done by the applied force on the crate is

$W_F=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here we have:

F = 73.5 N

d = 7.5 m

$\theta=0^{\circ}$ (the force is applied in the same direction as the displacement)

Therefore,

$W_F=(73.5)(7.5)(cos 0^{\circ})=551.3 J$

The work done by friction on the crate is:

$W_{F_f}=F_f d cos \theta$

where in this case:

$F_f=73.5 N$ is the magnitude of the force of friction

d = 7.5 m is the displacement of the crate

$\theta=180^{\circ}$ , because the displacement is forward and the force of friction is backward, so they are in opposite direction

Therefore, the work done by the force of friction is:

$W_{F_f}=(73.5)(7.5)(cos 180^{\circ})=-551.3 J$

To find the normal force, we analyze the situation of the force along the vertical direction.

We have two forces on the vertical direction:

- The normal force, N, upward

- The force of gravity, $mg$ , downward, where

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:

$N-mg=0\\N=mg=(50)(9.8)=490 N$

The work done by the normal force is:

$W_N=Nd cos \theta$

In this case, $\theta=90^{\circ}$ , since the normal force is perpendicular to the displacement of the crate; therefore, the work done is

$W_N=(490)(7.5)(cos 90^{\circ})=0$

The work done by the gravitational force is:

$W_g=F_g d cos \theta$

where:

$F_g=mg=(50)(9.8)=490 N$ is the gravitational force

d = 7.5 m is the displacement of the crate

$\theta=90^{\circ}$ is the angle between the direction of the gravitational force (downward) and the displacement (forward)

Therefore, the work done by gravity is

$W_g=(490)(7.5)(cos 90^{\circ})=0 J$

The total work done on the crate can be calculated by adding the work done by each force:

$W=W_F+W_{F_f}+W_N+W_g$

where we have:

$W_F=+551.3 J$ is the work done by the applied force

$W_{F_f}=-551.3 J$ is the work done by the frictional force

$W_N=0$ is the work done by the normal force

$W_g=0$ is the work done by the force of gravity

Substituting,

$W=+551.3+(-551.3)+0+0=0 J$

So, the total work is 0 J.

According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:

$W=\Delta E_K$

where

W is the work done on the crate

$\Delta E_K$ is the change in kinetic energy of the crate

In this problem, we have:

$W=0$ (total work done on the crate is zero)

Therefore, the change in kinetic energy of the crate is:

$\Delta E_K = W = 0$

5 0

3 years ago