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Natalka [10]
4 years ago
12

A ball is thrown horizontally from the top of a 41 m vertical cliff and lands 112 m from the base of the cliff. How fast is the

ball thrown horizontally from the top of the cliff?
Physics
1 answer:
yanalaym [24]4 years ago
4 0

Answer:

4.78 second

Explanation:

given data

vertical cliff = 41 m

height = 112 m

solution

we know here time taken to fall vertically from the cliff =  time taken to move horizontally   ..........................1

so we use here vertical component of ball

and that is accelerated motion with initial velocity = 0

so we can solve for it as

height = 0.5 ×  g ×  t²     ........................2

put here value

112 = 0.5 ×  9.8 ×  t²    

solve it we get

t²   = 22.857

t = 4.78 second

ball thrown horizontally from the top of the cliff in 4.78 second

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A spherical surface completely surrounds a collection of charges. Find the electric flux (with its sign) through the surface if
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Answer:

(a) 6.8 x 10^5 Nm^2/C

(b) 1.47 x 10^5 Nm^2/C

(c) 5.3 x 10^5 Nm^2/C

Explanation:

According to the Gauss's theorem

Electric flux = Charge enclosed / ∈0

(a) Charge enclosed = 6 x 10^-6 C

So, Electric flux = (6 x 10^-6) / (8.854 x 10^-12) = 6.8 x 10^5 Nm^2/C

(b) Charge enclosed = -1.3 x 10^-6 C

So, Electric flux = (1.3 x 10^-6) / (8.854 x 10^-12) = 1.47 x 10^5 Nm^2/C

(c) Charge enclosed = 6 x 10^-6 + (-1.3 x 10^-6) = 4.7 x 10^-6 C

So, Electric flux = (4.7 x 10^-6) / (8.854 x 10^-12) = 5.3 x 10^5 Nm^2/C

5 0
3 years ago
A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang
erma4kov [3.2K]

Given:

initial angular speed, \omega _{i} = 21.5 rad/s

final angular speed, \omega _{f} = 28.0 rad/s

time, t = 3.50 s

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Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

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Now, putting the given values in the above formula:

\alpha = \frac{28.0 - 21.5}{3.50}

\alpha = 1.86 m/s^{2}

Therefore, angular acceleration is:

\alpha = 1.86 m/s^{2}

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calculate the average velocity of a dancer who moves 5 Meters to the left of the stage over the course of 15 seconds
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A block rests on a ?at plate that executes vertical simple harmonic motion with a period of 0.58s.
SVETLANKA909090 [29]

Answer:

maximum amplitude  = 0.08 m

Explanation:

Given that

Time period T= 0.58 s

acceleration of gravity g= 9.8 m/s²

We know that time period of simple harmonic motion given as

T = 2π/ω

0.58 = 2π/ω

ω = 10.83rad/s

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Lets take amplitude = A

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A = 0.08m

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8 0
3 years ago
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