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gayaneshka [121]
2 years ago
15

Distinguishing Types of Mirrors Light is coming from the left; which is a concave mirror?

Physics
1 answer:
Snezhnost [94]2 years ago
7 0

A concave mirror is a converging mirror while a convex mirror is a diverging mirror.

<h3>Curved mirrors</h3>

Curved mirrors are different from plane mirrors because they have a center of curvature. The image formed by each type of mirror depends on the position of the object.

There are two types of curved mirrors;

  • Converging mirror
  • Diverging mirror

A concave mirror is a converging mirror while a convex mirror is a diverging mirror. The images are not provided here so we can not really say which is which.

Learn more about curved mirrors: brainly.com/question/8512677

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What condition occurs when the eyeball is too long?
Diano4ka-milaya [45]
Hello!

Myopia, also known as nearsightedness, occurs when the eyeball is too long. This condition is very common in the U.S and throughout the world. It can be corrected by wearing glasses or contact lenses.
7 0
3 years ago
Read 2 more answers
3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the
bulgar [2K]

Answer:

The wife have to sit at 0.46 L from the middle point of the seesaw.

Explanation:

We need to make a sketch of the seesaw and the loads acting over it.

And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.

A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.

The maximum momentum with respect to the pivot point (0) will be:

M=75 *\frac{L}{2}

Where L/2 is the half of the distance of the seesaw

Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.

7 0
3 years ago
If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the
Mazyrski [523]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

The potential of this system is  U=6.75*10^{-7}J

B

The electric potential at point p is V_p= -900V

C

The work required is  W= 9*10^{-7}J

D

The speed of the charge is  v=600m/s

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

            U = \frac{kq_1 q_2}{d}

where k is the electrostatic constant with a value of  k = 9*10^9 N m^2 /C^2

           q is the charge with a value of  q = 1*10^{-9}C

           d is the distance given as   d =5m

Now we are given that  q_1 = q and  q_2 = 3q and

Now substituting values

             U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}

                U=6.75*10^{-7}J

The electric potential at point P is mathematically obtained with the formula

             V_p = V_{-q} + V_{-3q}

I.e the potential at q_1 plus the potential at  q_2

Now potential at q_1 is mathematically represented as

                   V_{-q} = \frac{-kq}{s}

and the potential at q_2 is mathematically represented as

                        V_{-3q} = \frac{-3kq}{s}

Now substituting into formula for potential at  P

                  V_p = \frac{-kq}{s} + \frac{-3kq}{s}  = -\frac{4kq}{s}

                       = \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}

                      V_p= -900V

The Workdone to bring the third negative charge is mathematically evaluated as

                W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}

                                 = \frac{kq*q}{s} +   \frac{kq*3q}{s}

                                = \frac{4kq^2}{s}

                               = \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}

                              W= 9*10^{-7}J

From the Question are told that the charge q_3 would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically  represented as

                   \Delta U = W = \frac{1}{2} m_{q_3} v^2

                         9*10^{-7} = \frac{1}{2} m_{q_3} v^2

Where m_{q_3} = 5.0*10^{-12}kg

Now making v the subject we have

                 v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }

                     v=600m/s  

8 0
3 years ago
A biker goes up hill with a constant speed of 10km/h. Then he goes downhill with a constant speed of 50km/h. What's his average
Sladkaya [172]

This is not as simple as it looks.  

His average speed is NOT (10km/hr + 50km/hr)/2 = 30 km/hr.

You have to use the definition of speed:

Speed = (total distance covered) / (time to cover the distance).

Let's say the distance up (and down) the hill is 'd' .

Then the time it takes to go up the hill is (d/10) hours.

And the time it takes to come down the hill is (d/50) hours.

Total distance = 2d km

Total time = (d/10) + (d/50) = (5d/50) + (d/50) = 6d/50

Speed = distance/time = 2d/(6d/50) = 100d/6d

<em>Speed = </em>100/6 = <em>16-2/3 km/hr</em>

4 0
3 years ago
A 4 kg mass is in free fall. What is the velocity of the mass after 11 seconds
Ksju [112]
Velocity of the mass after 11 seconds = ( value of the gravitational acceleration) * ( time )
velocity = ( 9.81 m / s^2 ) ( 11)
velocity = 107.91 meters per second
5 0
3 years ago
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