I am attaching the missing figure
If we assume 100% efficiency, this means that the input power of the transformer must be equal to the output power
We divide output/input
240,000/12,000 = 20
This means that the transformer has a turns ratio of 1:20
<span>
ns/np = 20/1 = 20
The voltage is amplified 20 times by the transformer, but this also means that the current decreases by a factor of 20, so the total power remains the same.</span>
Answer:
897
Explanation:
Speed of the car, v = 126 km/h, converting to m/s, we have v = 35 m/s and
Radius of the curve, R = 150 mm = 0.15 m
The centripetal acceleration a(c) is given by the formula = v² / R so that
a(c) = 35² / 0.15
a(c) = 1225 / 0.15
a(c) = 8167 m/s²
The force that causes the acceleration is frictional force = µ m g, where
µ = coefficient of friction
m = the mass of the car and
g = acceleration due to gravity, 9.81
From Newton's law:
µ m g = m a(c) , so that
µ = a(c) / g
µ = 8167 / 9.81
µ = 897
Therefore, the coefficient of static friction must be as big as 897
Answer:
D. Copper
Explanation:
Copper has the best electrical conductivity of any metal, except silver. A good electrical conductivity is the same as a small electrical resistance.
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Answer:
The maximum height the pebble reaches is approximately;
A. 6.4 m
Explanation:
The question is with regards to projectile motion of an object
The given parameters are;
The initial velocity of the pebble, u = 19 m/s
The angle the projectile path of the pebble makes with the horizontal, θ = 36°
The maximum height of a projectile, 
, is given by the following equation;
Therefore, substituting the known values for the pebble, we have;
Therefore, the maximum height of the pebble projectile, ≈ 6.4 m.
Answer:
<em>-6,329.5Joules</em>
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Explanation:
Complete question:
Sally applies a horizontal force of 462N with a rope to drag a wooden crate across a floor with a constant speed the rope tied to the crate is pulled at an angle of 56.0degree and sally moves the crate 24.5m. What work is done by the floor through the force of friction between the floor and crate.
Work done = Fd cos theta
F is the horizontal force
d is the distance covered
theta angle of inclination
Substituting into the formula
Workdone = 462(24.5)cos 56
Workdone = 11,319(0.5592)
Workdone = 6,329.5Joules
Hence the workdone by sally is 6,329.5Joules
<em>The work done by friction will be opposite the work done by sally, hence work done by the floor through force of friction between the floor and the crate is -6,329.5Joules</em>
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