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Mila [183]
4 years ago
15

A battery is connected to a light bulb in a circuit. The voltage of the battery is 1.5 V(volts). The light bulb has a resistance

of 1.5 Ohms. What is the current in the light bulb?
Use the formula i= v/r to solve this problem.
Physics
1 answer:
trasher [3.6K]4 years ago
5 0
Voltage = 1.5V
Resistance = 1.5 ohm

Current = V/R = 1.5/1.5 = 1A
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A basebal (radius = .036 m, mas = .145kg) is droped from rest at the top of the Empire State Building (height = 1250ft). Calcula
lora16 [44]

Answer:

a) the initial potential energy = 541.95J

b) the final kinetic energy = 87.991 J

Explanation:

<u>Step 1</u>: Data given

The ball has a mass of 0.145 kg and is at a height of 381m

<u>Step 2</u>: Calculate potential energy

The potential energy = m * g * h

with m = the mass of the ball = 0.145 kg

with g = Gravitational acceleration = 9.81 m/s²

with h = the height of the building = 381m

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FD = 1/2ρCDAv²

⇒ with FD = the drag force = the force component in the direction of the flow velocity

⇒ with ρ = density of the fluid (air in our case: ρ≈1.1839 Kg/m3 at 1 atm and 25 °C)

⇒ with v = velocity of the ball

⇒ with A = reference area, which in our case is just the cross sectional area of the ball: A=πr2

⇒ with CD is the drag coefficient - a dimensionless coefficient, that in the case of a sphere, CD=0.47

Following Newton's second law:

ΣFy = may = -mg +Dv²

Here is D=1/2ρCDA ( for convenience) = 0.001172

The terminal speed we can define as the speed of the ball where ay = 0

Therefore: -mg + Dvt² = 0

⇒vt = √(mg/D)

vt = √(0.145 * 9.8 / 0.001172) ≈ 34.837

Uinitial=mgh≈541.951 J (see first question)

Kfinal=1/2*mvt²=(m²g)/2D ≈ 87.991 J

The final kinetics energy is 87.991 J

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