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iragen [17]
1 year ago
12

6.A string of 100 Christmas lights requires 4 Volts and 2 Amps for a single bulb to light up. (a) When one light on the sting go

es out the whole string of lights no longer turns on. What type of circuit are these lights? (b) What is the total voltage, current, and resistance of this circuit to light up all 100 bulbs? (c) When one light goes out on the string all other lights still turn on. What type of circuit are these lights? (d) What is the total voltage, current, and resistance of this circuit to light up all 100 bulbs?

Physics
1 answer:
Molodets [167]1 year ago
3 0

Answer:

Explanation:

From the information given,

V = 4 volts

A = 2 amps

a)

In the first instance, one light on the sting goes out and the whole string of lights no longer turns on. This means that the circuit is a series circuit.

b) Total voltage = 4 x 100 = 400 V

The current passing through each bulb is the same. Thus

Total Current = 2 Amps

Recall, V = IR

R = V/I

Thus,

Resistance = 400/2

Resistance = 200 ohms

c) In this case, one light goes out on the string all other lights still turn on. This means that the circuit is a parallel circuit

d)

The voltage is the same

Total voltage = 4 volts

Total current = 2 x 100 = 200 amps

Total resistance = 4/200

Total resistance = 0.02 ohms

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The period of the orbit would increase as well

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We can answer this question by applying Kepler's third law, which states that:

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Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
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Answer:

45.6m

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The equation for the position y of an object in free fall is:

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With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

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Combining equation 1 and 4:

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3 years ago
You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i
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The general formula to calculate the work is:

W=Fd \cos \theta

where F is the force, d is the displacement of the couch, and \theta is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.


(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so \theta=0^{\circ} and \cos \theta=1, therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

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(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore \theta=90^{\circ} and \cos \theta=0: this means

W=0.


(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

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