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RUDIKE [14]
3 years ago
12

From what height should you drop a ball if you want it to hit the ground in exactly 3 seconds?

Physics
1 answer:
Trava [24]3 years ago
5 0

Answer:

44.1 m

Explanation:

Given:

v₀ = 0 m/s

a = -9.8 m/s²

t = 3 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (3 s) + ½ (-9.8 m/s²) (3 s)²

Δy = -44.1 m

The ball should be dropped from 44.1 meters.

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Neporo4naja [7]

The correct answer is y=-2x+(1/2)

y = f'(x)· x + c

Y = -2x + C

1 = -2x π/4 + C

=) C = I + π/2

y=-2x+(1/2)  is the first-degree polynomial.

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1 year ago
What can be found by counting the number of troughs per second in a wave diagram? amplitude direction frequency speed wavelength
Nutka1998 [239]
The period T is time it takes for one complete cycle or from "trough to trough" so the reverse is trough per sec = 1/T = frequency
6 0
3 years ago
Read 2 more answers
When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point
vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity \rightarrow 20m/s

\lambda = Wavelength \rightarrow 35*10^{-2}m

Therefore the frequency would be given as

f = \frac{20}{35*10^{-2}}

f = 57.14Hz

The frequency is directly proportional to the angular velocity therefore

\omega = 2\pi f

\omega = 2\pi *57.14

\omega = 359.03rad/s

Now the maximum speed from the simple harmonic movement is given by

V_{max} = A\omega

Where

A = Amplitude

Then replacing,

V_{max} = (1*10^{-2})(359.03)

V_{max} = 3.59m/s

Therefore the maximum speed of a point on the string is 3.59m/s

8 0
3 years ago
A rocky space object of varying size
kow [346]
That would be an asteroid
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3 years ago
An automobile accelerates 1.77 m/s2 over 6.00 s to reach freeway speed at the end of an
GaryK [48]

Answer:

Startinfg speed is 13.82 m/s

Explanation:

Use equation for realtion between start and final speed :

Vf=Vs+a*t

Vf-final speed

Vs-start speed

Vf=24.44m/s

a=1.77m/s²(acceleration)

t=6.00s(Time)

Vf=Vs+a*t

Vs=a*t-Vf

Vs=1.77m/s²*6s-24.44m/s

Vs=-13.82m/s

6 0
3 years ago
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