NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question
Answer:
Power = 2702.56 W
Explanation:
Let the power consumed be P
Energy expended = E = mgh
height, h = 5 m
E = 80 * 9.8 * 5
E = 3920 J
To calculate the time, t
From F = ma
F = 900 N
900 = 80 a
a = 900/80
a = 11.25 m/s²
From the equation of motion, 
The drill head starts from rest, u = 0 m/s
Power, P = E/t
P = 3920/0.0.943
P = 4157.79 W
But Efficiency, E = 0.65
P = 0.65 * 4157.79
Power = 2702.56 W
Answer:
Alternating
Explanation:
It is alternating because it is easy to distribute long distance.
Direct current is found in batteries and have large voltage drop over distance.
Answer:
8.57 Hz
Explanation:
From the question given above, the following data were obtained:
Wavelength (λ) = 3.5 m
Velocity (v) = 30 m/s
Frequency (f) =?
The velocity, wavelength and frequency of a wave are related according to the equation:
Velocity = wavelength × frequency
v = λ × f
With the above formula, we can simply obtain the frequency of the wave as follow:
Wavelength (λ) = 3.5 m
Velocity (v) = 30 m/s
Frequency (f) =?
v = λ × f
30 = 3.5 × f
Divide both side by 3.5
f = 30 / 3.5
f = 8.57 Hz
Thus, the frequency of the wave is 8.57 Hz
<span>R = rate of flow = 0.370 L/s
H = height = 2.9 m
T= time = 3.9 s
V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2
G value = 9.8 m/s2
Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N
Wa = weight of accumulated water after 3.9 s
Fi = force of impact of water on the bucket
S = reading on the scale = Wa + Wb + Fi
mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg
Therefore, Wa = 1.443 x 9.8 = 14.1414 N
Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt)
= 0.37 x 7.539 = 2.78943 N
S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
Answer:
(a) Vf = 128 ft/s
(b) K.E = 122.8 Btu
Explanation:
(a)
In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = 32.2 ft/s²
h = height = 253 ft
Vf = Final Velocity = ?
Vi = Initial Velocity = 10 ft/s
Therefore,
(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²
16293.2 ft²/s² + 100 ft²/s² = Vf²
Vf = √(16393.2 ft²/s²)
<u>Vf = 128 ft/s</u>
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(b)
The kinetic energy of the object before it hits the surface of earth is given by:
K.E = (0.5)(m)(Vf)²
where,
m = mass of object = 375 lb
K.E = Kinetic energy of object before it strikes the surface of earth = ?
Therefore,
K.E = (0.5)(375 lb)(128 ft/s)²
K.E = 3073725 lb.ft²/s²
Now, converting this to Btu:
K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)
<u>K.E = 122.8 Btu</u>
