Resistance of a lightbulb is 240.8 ohm
Given:
Power = 120 W
Frequency = 60.0-Hz
Voltage = 170 V
To Find:
Resistance of a lightbulb
Solution: The electrical resistance of a circuit is the ratio between the voltage applied to the current flowing through it. Rearranging the above relation, R = V I. The unit of electrical resistance is ohms.
The resistance of the light bulb is given by the formula, R=V²/P
R = (170)^2/120
R = 240.8 ohm
Hence, the resistance of light bulb is R = 240.8 ohm
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Explanation:
v = u + at
25 = 75 + a * 5
25 = 75 + 5 a
a = 25 - 75/5
a = -50/5
a = -10m/sec^2
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Answer:
w = 1,066 rad / s
Explanation:
For this exercise we use Newton's second law
F = m a
the centripetal acceleration is
a = w² r
indicate that the force is the mass of the body times the acceleration
F = m 0.58g = m 0.58 9.8
F = 5.684 m
we substitute
5.684 m = m w² r
w =
To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m
w =
w = 1,066 rad / s
Answer:
Explanation:
<u>The pressure of a Fluid</u>
A fluid of density exerts pressure at a distance y (deep) given by
Where g is the acceleration of gravity or
This formula computes the pressure assuming the initial pressure is 0 at fluid (water in this case) level.
Knowing the measured pressure, we can know how deep the diver went by solving the equation for y
Let's plug in the given values
Thus