For this, we must know the density of lead. This is 11.36 grams per cubic centimeter. The mass present is 408 grams. Therefore, the volume occupied will be:
408 / 11.36 = 35.9 cubic centimeters. This is the volume of water that will be displaced.
Now, we know that one cubic centimeter is equivalent to one mL. Therefore,
35.9 ml of water will be displaced.
Given Information:
Voltage of circuit A = Va = 208 Volts
Current of circuit A = Ia = 40 Amps
Voltage of circuit B = Vb = 120 Volts
Current of circuit B = Ib = 20 Amps
Required Information:
Ratio of power = Pa/Pb = ?
Answer:
Ratio of power = Pa/Pb = 52/15
Explanation:
Power can be calculated using Ohm's law
P = VI
Where V is the voltage and I is the current flowing in the circuit.
The power delivered by circuit A is
Pa = Va*Ia
Pa = 208*40
Pa = 8320 Watts
The power delivered by circuit B is
Pb = Vb*Ib
Pb = 120*20
Pb = 2400 Watts
Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is
Pa/Pb = 8320/2400
Pa/Pb = 52/15
Answer:
23 m/s downward
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<em>Taking the downward direction as positive</em>
<u>We are given:</u>
Initial velocity of the marble (u) = 0 m/s
Time interval (t) = 2.3 seconds
Final velocity (v) = x m/s
<u>Solving for the Final velocity:</u>
<u>Acceleration of the Marble:</u>
We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s
<u>Final velocity:</u>
v = u + at [First equation of motion]
x = 0 + (10)(2.3) [replacing the given values]
x = 23 m/s
Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction
Answer:
11.8 m/s
Explanation:
At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).
Sum of forces in the centripetal direction:
∑F = ma
mg − N = m v²/r
At the maximum speed, the normal force is 0.
mg = m v²/r
g = v²/r
v = √(gr)
v = √(9.8 m/s² × 14.2 m)
v = 11.8 m/s
The parallel component is given by
F=180cos(25)=163.14N
