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3 years ago

11

Sal connected four light bulbs in a parallel circuit. If Sal adds another light bulb, what is true about the current in each bul

b?
A) The current decreased when a fourth bulb was added.

B) The current in the four bulbs is much less than the current in three bulbs.

C) The current in each bulb remained the same even though another bulb was added.

D) The current in the fourth bulb is less than the current int he other three bulbs.

2 answers:

Zinaida [17]3 years ago

4 0

The answer is C) The current in each bulb remained the same even though another bulb was added.

kolbaska11 [484]3 years ago

4 0

Since all the bulbs are connected in parallel with each other

so here voltage across each bulb will remains the same

so here we will have

$V = i R$

now all bulbs are identical and we have same voltage source across all the bulbs

so we will have current in each bulb given by same equation

$i = \frac{V}{R}$

now we can say that correct answer will be

C) The current in each bulb remained the same even though another bulb was added.

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What causes a satellite to move tangentially to a circular path?

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Explanation : We know that two forces acting on the satellite.

The speed of the satellite and the force of the gravity on the satellite then the result , the satellite moves around the earth in a circular motion.

A satellite to move tangentially to a circular path, while revolving around the circular path because the centripetal force is acting on it.

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3 years ago

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A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

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Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

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3 years ago

Select the correct answer

devlian [24]

Explanation:

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3 years ago

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A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as

Taya2010 [7]

Answer:

$w = - 508.53$ joules

$q = - 3091.47$ joules

Explanation:

Let us convert the time in hours into seconds

$0.010* 3600\\= 36$

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$\delta E = p * \delta t$

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

$\delta E = - 100 * 36\\$

$\delta E = - 3600$ Joules

Amount of work done by the system

$w = - P * \delta V$

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

$w = - 1 * ( 5.92 -0.90)\\$

$w = -5.02$ liter-atmospheres

Work done in Joules

$- 5.02 * 101.3\\$ $= 508.53$ Joules

$q = \delta E - w\\$

Substituting the given values we get -

$q = - 3600 - (-508.53)\\q = - 3091.47$

Thus

$w = - 508.53$ joules

$q = - 3091.47$ joules

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3 years ago

Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th

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Answer:

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3 years ago