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2 days ago

Two 8.0 Ω lightbulbs are connected in a 12 V series circuit. What is the power of both glowing bulbs?

Physics

1 answer:

inna [2.7K]2 days ago

7 0

Answer:

18 W

Explanation:

Using the formula,

P = V²/R.................. Equation 1

In this formula, P represents the power of both bulbs, V is the voltage, and R is the total resistance.

Given that this is a series circuit,

We have

R = R1 + R2............. Equation 2

where R1 signifies the resistance of the first bulb and R2 is the resistance of the second bulb.

We know R1 = R2 = 8 Ω

Now substituting into equation 1 gives us

R = 8 + 8

R = 16 Ω

It's also given that V = 12 V

Substituting into equation 1, we find

P = 12²/16

P = 144/16

P = 9 W

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Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

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Response:

Magnitude of the electrostatic force acting on the +32 µC charge, $F_{net} = 12 N$

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

$F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }$

$F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N$

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).

$F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }$

$F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N$

The resultant electrostatic force on the 32 µC charge is $F_{net} = |F_{2} + F_{3}|$

$F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N$

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1 month ago

Rosa studies the position-time graph of two race cars. A graph titled Position versus Time shows time in hours on the x axis, nu

Sav [2813]

Answer:

B. Truck X was ahead, not truck Y.

Explanation:

Let's analyze the information provided.

Truck X moved from the point (0,20) to (2.8,50). This indicates that it began at the 20th kilometer and reached 50 km in 2.8 hours. Thus, its speed is v1 = (s2 - s1) / t

v1 = (50 - 20) / 2.8

v1 = 10.7 km/h

Given that it started from the 20th km, it indeed had a head start. Since the line on the graph is linear, this shows its speed was constant without any change in direction.

On the other hand, Truck Y's movement went from the origin (0,0) to (5,20), meaning it took 5 hours to travel 20 km, resulting in a speed of v2 = 20 / 5

v2 = 4 km/h

Again, the straightness of its graph line signifies it maintained a constant speed in a single direction.

Thus, it is evident that Rosa erred in her assumption that Truck Y had a head start.

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1 month ago

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A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav

kicyunya [2911]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

The locomotive's acceleration is 1.6 $m/s^2$

The duration taken to pass the crossing is 2.4 seconds.

We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 $m/s^2$ .

32 = 0 + 1.6 * t

t = 20 seconds.

Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

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1 month ago

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kicyunya [2911]

Answer:

$\Delta T = 0.81 ^oC$

Explanation:

According to the principle of energy conservation

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$\frac{1}{2}mv^2 = ms\Delta T$

Next, divide both sides by the object's mass

$\frac{1}{2}v^2 = s\Delta T$

the resulting temperature change is expressed as

$\Delta T = \frac{v^2}{2s}$

$\Delta T = \frac{25^2}{2\times 387}$

$\Delta T = 0.81^oC$

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The appropriate choice is C.

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