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Ainat [17]
2 years ago
10

1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.

Physics
1 answer:
Mashutka [201]2 years ago
4 0

The position of the object is = -68cm

The magnification of the mirror= 0.3

<h3>Calculation of object distance</h3>

The image distance = 20.5cm

The focal length= R/2 = 31.5/2= 15.75

The object distance= ?

Using the lens formula,1/f = 1/v-1/u

1/u = 1/v- 1/f

1/u = 1/20.5 - 1/15.75

1/u = 0.0489- 0.0635

1/u = -0.0146

u = -68cm

The magnification of the mirror is image size/object size

= 20.5cm/-68cm

= 0.3

Learn more about lens here:

brainly.com/question/9757866

#SPJ1

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10 points
grandymaker [24]
The answer is B tell me if I am wrong.
6 0
2 years ago
A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?
denis23 [38]

Answer:

Workdone = 465766038 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

7 0
2 years ago
Explain what major mistake producers made when editing the battle scenes for the movie Star Wars
Artyom0805 [142]

Answer: Air

Explanation:

Roffns dn b bdd

4 0
2 years ago
A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full
AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is  is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

A=\pi r^2

Put the value into the formula

A=\pi\times21^2

A=441\pi\ feet^2

Thickness of slice t=\Delta h\ ft

The volume is,

V=(441\pi\times\Delta h)\ ft^3

We need to calculate the force

Using formula of force

F=W\times V

Where, W = water weight

V = volume

Put the value into the formula

F=62.5\times(441\pi\times\Delta h)

F=27562.5\pi\times\Delta h\ lbs

We need to calculate the work done

Using formula of work done

W=F\times d

Put the value into the formula

W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

W=\int_{0}^{h}{W}

Put the value into the formula

W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh

W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}

W=27562.5\pi(10\times10-\dfrac{100}{2}-0)

W=4329507.37572

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

W=\int_{0}^{h}{W}

Put the value into the formula

W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh

W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}

W=27562.5\pi(12\times10-\dfrac{100}{2}-0)

W=1929375\pi

W=6061310.32\ foot- pound

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is  is 6061310.32 foot-pound.

8 0
3 years ago
What may result when the energy that builds up at plate boundaries is released because the plates suddenly overcome the force of
Arlecino [84]
It has to be Earthquake
4 0
2 years ago
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