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3 years ago

1. What will happen to the brightness of the light bulb if the switch in this circuit is suddenly closed?

Physics

1 answer:

bazaltina [42]3 years ago

6 0

Answer:

There is no closed-loop path for the current to flow through the circuit.When the switch is closed,the light bulb operates since the current flows through the circuit.The bulb glows at its full brightness since its receives its full 120 volts and has the design current flow.

Explanation:

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A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga

BartSMP [9]

Answer:

Total work done in expansion will be $3.60\times 10^5J$

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm $=1.01\times 10^5Pa$

So 2.10 atm $=2.10\times 1.01\times 10^5=2.121\times 10^5Pa$

Volume is increases from 3370 liter to 5.40 liter

So initial volume $V_1=3.70liter$

And final volume $V_2=5.40liter$

So change in volume $dV=5.40-3.70=1.70liter$

For isobaric process work done is equal to $W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J$

So total work done in expansion will be $3.60\times 10^5J$

5 0

3 years ago

Usain Bolt accelerates at a rate of 3.7

MissTica

Answer:

Explanation:

I think that you to run more than 12 miles

4 0

4 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

6 0

3 years ago

The planet in our solar system whose orbit actually brings it inside the orbit of another planet is

Ber [7]

Answer:

Pluto

Explanation:

In our solar system, we have several planet. Pluto is one of the. Pluto is a planet that is highly oval shaped orbit and eccentric that brings it inside the another orbit. It get inside the orbit of Neptune. Sometimes even Neptune get far away from sun in comparison to the dwarf planet Pluto.

It is very strange happening in the world of planet. it occurs in the year of 1979 and 1999. But Pluto never ever crashed into Neptune. It happen because Neptune takes every three lapse that takes around the sun but Pluto makes only two lapse. This happening prevents two bodies from clashes.

7 0

4 years ago

You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta

Arte-miy333 [17]

Answer:

1) 3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes: $\frac{Distance}{Speed}= \frac{22}{6} = 3.66$

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes: $Speed * Time = 34 * 3.66 = 124.44$

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : $s = u*t + \frac{1}{2}(a * t^2)$

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

$22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)$

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1 = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.

4 0

3 years ago