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2 years ago

6

A cricket ball is dropped from a height of 20 m. Calculate: a) the speed of the ball. b) the time it takes to fall through this

height

1 answer:

AysviL [449]2 years ago

7 0

Answer:

Initial velocity is 0 .. ( since it is just dropped)

now using V²= u² +2gh

=> (Vfinal)² = 0+2*10*20

=> v² = 20*20 = 400

=> v = √400 = 20m/s

for time taken

use V = u+gt

=> t = V/g = 20/10 = 2sec

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Water flows over a section of Niagara Falls at rate of 1.1×10^6 kg/s and falls 49.4 m. How much power is generated by the fallin

DerKrebs [107]

Answer:

<em>The power generated is = 5.33×10⁸ Watt. </em>

Explanation:

Power: Power can be defined as the time rate of doing work. The S.I unit of power is <em>Watt(W).</em>

<em>Mathematically,</em>

<em>Power (P) = Work done/time or Energy/time</em>

P = mgh/t............................... Equation 1

P = δgh............................. Equation 2

Where δ = fall rate, g = acceleration due to gravity, h = height.

<em>Given: </em>δ = 1.1×10⁶ kg/s, h = 49.4 m g = 9.81 m/s²

Substituting these values into equation 2

P = 1.1×10⁶×49.4×9.81

P = 533.08×10⁶

<em>P = 5.33×10⁸ Watt.</em>

<em>Thus the power generated is = 5.33×10⁸ Watt. </em>

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3 years ago

In the context of depth perception, which of the following is a monocular cue?

olga55 [171]

Hello there.

<span>In the context of depth perception, which of the following is a monocular cue?

</span><span>(C) Convergence
</span>

8 0

3 years ago

The distance between two slits is 1.50 *10-5 m. A beam of coherent light of wavelength 600 nm illuminates these slits, and the d

Fed [463]

Answer: y = 2.4×10^-6m or y= 2.4μm

Explanation: The formulae for the distance between the central bright fringe to any other fringe in pattern is given as

y = R×mλ/d

Where y = distance between nth fringe and Central bright spot fringe.

m = position of fringe = 4

λ = wavelength of light= 600nm = 600×10^-9 m

d = distance between slits = 1.50×10^-5m

R = distance between slit and screen = 2m

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y = 4800×10^-9/2

y = 2400 × 10^-9

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3 years ago

A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h

boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α

where, α is the angular acceleration

α = a / R

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from (1), (2) and (3)

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a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

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3 years ago

What is an Amplitude

dangina [55]

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3 years ago

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