Question 15)

In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of th

Charra [1.4K]

Answer:

1960.32306 kg/s

Explanation:

m = Mass of water = 1 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height from which the water will fall

Potential Energy

$PE=mgh\\\Rightarrow PE=1\times 9.81\times 52\\\Rightarrow PE=510.12\ J$

One megawatts of power is required

So, flow rate

$\dotm=\frac{P}{PE}\\\Rightarrow \dotm=\frac{1\times 10^6}{510.12}\\\Rightarrow \dotm=1960.32306\ kg/s$

1960.32306 kg/s is required to produce a megawatt of power

7 0

3 years ago

The density of air under normal conditions is about 3 1.2 kg/ m . For a wind speed of 10 m/s, find (a) the actual power carried

dusya [7]

The density’ is 5.46

3 0

3 years ago

A student is flying west on a school trip from Winnipeg to Calgary in a jet that has an air velocity of 792 km/h.The direction t

Akimi4 [234]

Answer:

The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h[N] is 4.5° S of W

Explanation:

The given parameters are;

Velocity of Jet = 792 km/h

Direction of jet velocity = West

Velocity of wind = 62.0 km/h

Direction of wind velocity = North

Therefore, the jet has to have a component of 62.0 km/h South of West to compensate for the wind velocity

The direction of the plane, θ° South of West (S of W) to compensate for the wind is given as follows;

$Tan \left (\theta \right )= \dfrac{62}{792} = \dfrac{31}{396}$

Therefore;

$\theta = tan^{-1}\left (\dfrac{31}{396} \right ) = 4.476^{\circ} \approx 4.5^{\circ}$

The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h[N] = 4.5° S of W.

6 0

4 years ago

A model rocket blasts off and moves upward with an acceleration of 12 m/s2 until it reaches a height of 26 m. at that height, it

Diano4ka-milaya [45]

initial acceleration of rocket is given as

a = 12 m/s^2

h = 26 m

now we can use kinematics to find its speed

$v_f^2 - v_i^2 = 2 a h$

$v_f^2 - 0 = 2 * 12*26$

$v_f = 24.98 m/s$

now after this it will be under free fall

so now again using kinematics

$v_f = 0$

at maximum height

$v_f^2 - v_i^2 = 2 a s$

$0 - 24.98^2 = 2 * (-9.8)* h$

$h = 31.8 m$

total height from the ground = 31.8 + 26 = 57.8 m

Part b)

now after reaching highest height it will fall to ground

So in order to find the speed we can use kinematics again

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2*9.8*57.8$

$v_f = 33.67 m/s$

Part c)

first rocket accelerate to reach height 26 meter and speed becomes 24.98 m/s

now we have

$v_f - v_i = a t$

$24.98 - 0 = 12*t_1$

$t_1 = 2.1 s$

after this it will reach to highest point and final speed becomes zero

$v_f - v_i = at$

$0 - 24.98 = -9.8 * t$

$t_2 = 2.55 s$

now from this it will fall back to ground and reach to final speed 33.67 m/s

now we have

$v_f - v_i = at$

$33.67 - 0 = 9.8 * t$

$t_3 = 3.44 s$

so total time is given as

<em>t = 3.44 + 2.55 + 2.1 = 8.1 s</em>

5 0

3 years ago

In 2-4 sentences, summarize what you know about EM waves.

MaRussiya [10]

Answer:

In physics, electromagnetic radiation refers to the waves of the electromagnetic field, propagating through space, carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, light, ultraviolet, X-rays, and gamma rays.

8 0

3 years ago

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