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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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A ball is dropped from a cliff. determine how far the ball fell after 7.5 seconds
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Answer:

The ball fell 275.625 meters after 7.5 seconds

Explanation:

<u>Free fall </u>

If an object is left on free air (no friction), it describes an accelerated motion in the vertical direction, powered exclusively by the acceleration of gravity. The formulas needed to compute the different magnitudes involved are

V_f=gt

\displaystyle y=\frac{gt^2}{2}

Where V_f is the final speed of the object in free fall, assumed positive downwards, t is the time elapsed since the release and y is the vertical distance traveled by the object

The ball was dropped from a cliff. We need to calculate the vertical distance the ball went down in t=7.5 seconds. We'll use the formula

\displaystyle y=\frac{gt^2}{2}

\displaystyle y=\frac{(9.8)(7.5)^2}{2}

Y=275.625\ m

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3 years ago
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a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres​
Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

           l=4 \times 9=36 m

7 0
3 years ago
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xxMikexx [17]

Answer:

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