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Sergio039 [100]
4 years ago
6

A 60 W light bulb is powered by a connection to a wall outlet with 120 V across the plug terminals. a) What is the current passi

ng thru this resistor?
Physics
2 answers:
Oxana [17]4 years ago
7 0

Answer:

The current through the resistor is 0.5 A

Explanation:

Given;

power of the light bulb = 60 W

voltage in the wall outlet across the plug terminals = 120 V

power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.

power = voltage x current

Current = \frac{power}{voltage} = \frac{60}{120} = 0.5 A

Therefore, for a  60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A

bulgar [2K]4 years ago
5 0

Answer:

0.5 A

Explanation:

Current: The can be defined as the rate of flow of charge around a circuit. The S.I unit of current is Ampere(A)

Using,

P = VI..................... Equation 1

Where P = Power rating of the bulb, V = Voltage of the wall outlet, I = Current passing through the resistor.

Make I the subject of the equation

I = P/V................. Equation 2

Given: P = 60 W, V = 120 V.

Substitute into equation 2

I = 60/120

I = 0.5 A.

Hence the  current passing through the resistor = 0.5 A

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Answer:

269 m

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Explanation:

Part 1

First, find the time it takes for the package to land.  Take the upward direction to be positive.

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v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

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Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

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Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

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3 years ago
Why do I need to use the Unit Circle in Physics? And how do I use it?
stiks02 [169]

Answer:

The unit circle helps in making so many calculations and equations easy.

Explanation:

You cannot separate the knowledge of trigonometry to application in equations to physics. The unit circle is known to have a radius of one. This means that the distance from the centre of the circle, regardless of the unit of measurement, to any point of the edge of the circle is 1. Since the unit circle is very helpful in trigonometry, and trigonometry in turn is the projection of triangles and angles that is very crucial in the calculation of momentum, velocity and other factors of physics, the importance of the unit circle cannot be overemphasized in physics.

6 0
3 years ago
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