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ddd [48]
3 years ago
8

HURRY ! the BEST I will give the brainliest​

Physics
1 answer:
Oksanka [162]3 years ago
7 0

~Hello There!~

v^{2} =u^{2} +2as

a = \frac{v-u}{t}

40^{2} =2(\frac{40-0}{8})s

1600 = 10s

s = 160

s = distance traveled til car stopped.

180 - 160 = 20

The distance of stopped car from obstacle is 20m

Hope This Helps You!

Good Luck :)

Have A Great Day ^_^

- Hannah ❤

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Answer:

v = 9.936 m/s

Explanation:

given,

height of cliff = 40 m

speed of sound = 343 m/s

assuming that time to reach the sound to the player = 3 s

now,

time taken to fall of ball

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 40}{9.8}}

t = 2.857 s

distance

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d = v x 2.875

time traveled by the sound before reaching the player

t_0 = t - t_{fall}

t_0 = 3 - 2.875

t_0 = 0.143 s

distance traveled by the wave in this time'

r = 0.143 x 343

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7 0
3 years ago
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Dmitrij [34]

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3 0
3 years ago
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A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a
Verizon [17]

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

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(a) lens is a convex lens with

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object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

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⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0
3 years ago
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Burka [1]

To develop this problem it is necessary to apply the concepts related to the uni-axial deflection of bodies.

From the expression of Hooke's law we have to

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And substituting P/A for stress and \delta/L for strain gives that

\frac{P}{A} = E\frac{\delta}{L}

Where,

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\delta = \frac{PL}{AE}

If we want to calculate the deformation per unit area then we can also rewrite the equation as

\frac{delta}{L} = \frac{P}{AE}

Replacing with our values we have to

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