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shtirl [24]
3 years ago
9

A 60-watt light bulb has a voltage of 120 bolts applied across it and a current of 0.5 amperes flows through the bulb. What is t

he resistance of the light bulb?
Physics
1 answer:
Andre45 [30]3 years ago
7 0

Answer:

240 ohms

Explanation:

From Ohms law we deduce that V=IR and making R the subject of the formula then R=V/I where R is resistance, I is current and V is coltage across. Substituting 120 V for V and 0.5 A for A then

R=120/0.5=240 Ohms

Alternatively, resistance is equal to voltage squared divided by watts hence \frac {120^{2}}{60}=240

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The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electro
shusha [124]

Answer:

1.6\times 10^{-18} C  

Explanation:

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3 years ago
Suppose a car of mass m is moving at a constant speed v of
SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

\arcsin(v^2/(gR))

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In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

w_r = mg\sin(\theta)

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g

Also, the car is making a circular motion:

a = \frac{v^2}{R}

Combining the equations:

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3 years ago
A 0.10 newton spring toy with a spring constant of 160 newtons per meter is compressed 0.05 meter before it is launched. When re
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Answer:

(1) V = 0.2 J (2) 0.05J

Explanation:

Solution

Given that:

K = 160 N/m

x = 0.05 m

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(1) we solve for the  initial potential energy stored

Thus,

V = 1/2 kx² = 0.5 * 160 * (0.05)²

Therefore V = 0.2 J

(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.

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8 0
3 years ago
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