GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity.
U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J
U = 686.7J
Hope this helps!
The Bernoulli's principle is <span>the principle in hydrodynamics that an increase in the velocity of a stream of fluid results in a decrease in pressure.</span>
Answer:
Density of liquid = 4730 kg/m³
Atmospheric pressure on planet X = 8401.7 N/m²
Explanation:
Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm
So, ρgh = ρ₁gh₁
ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³
The atmospheric pressure on planet X
P = ρg₁h₃ g₁ = g/4 and h₃ = 725 mm = 0.725 m
on planet X
P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²
The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.
<h3>What is power of the circuit?</h3>
The power of the bulb or any resistor is equal to the product of voltage and current flowing through it.
P = VI
Circuit A has bulbs in series while the circuit B has bulbs in parallel.
When bulb 3 added to circuit A, the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.
The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.
Thus, the last option is correct.
Learn more about power.
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