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3 years ago

A ball is dropped from the top of a 77 m building. With what speed does the ball hit the ground? _________ m/s

Physics

1 answer:

vitfil [10]3 years ago

3 0

Answer:

38.87 m/s

Explanation:

Given that the ball is dropped from a height = 77 m

u = 0 m/s

s = 77 m

a = g = 9.81 m/s²

Applying the expression as:

$v^2-u^2=2as$

Applying values as:

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 77+0^2}\\\Rightarrow v=38.87\ m/s$

<u>The speed with which the ball hit the ground = 38.87 m/s</u>

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For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
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which becomes
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$A_1=0.030 m^2$ is the area of the first piston
$F_2$ is the force on the second piston
$A_2=0.015 m^2$ is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
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3 years ago

Read 2 more answers

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emmainna [20.7K]

Answer:

64.945 miles per hour

Explanation:

Since the frequency of sound heard is higher than actual frequency, the ambulance is moving towards you!

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = 8.61 KHz

f₀ = real frequency of sound = 7.87 KHz

v₀ = velocity at which the sound source is moving towards the reference point = ?

v = velocity of sound waves = 343 m/s

8.61 = 7.87/(1 - (v₀/v))

1 - (v₀/343) = 0.9141

v₀/343 = 1 - 0.9141 = 0.0859

v₀ = 343 × 0.0859 = 29.48 m/s = 64.945 miles per hour

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3 years ago

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One more thing—you posted this in the physics section rather than biology.

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3 years ago

A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the

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<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal

it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
--------------------------------------...
let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
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=====================
t = HALF the time of flight
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1 = u^2 sin p cos p/g
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5.42 * u cos p = 9.8
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check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
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so it will able to cross</span>

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3 years ago

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stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$