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Rudik [331]
2 years ago
11

A particular laser developed in 1995 at the University of Rochester, in New York, produced a beam of light that lasted for about

one-billionth of a second. The power output of this beam was 6.0 x 10^ 13 W. Assume that all of the electrical power was converted into light and that 8.0 x 10^6 A of current was needed to produce this beam. How large was the voltage that produced the current?
Physics
1 answer:
NISA [10]2 years ago
4 0

Power is the energy per unit time. When the current flow is needed to produce the beam. The voltage that produced the current is 7.5 x 10⁶ Volts.

<h3>What is power?</h3>

When voltage difference is created between the end of the conductor, the the power produced is the product of voltage and current.

P =VI

The power output of the beam is 6.0 x 10¹³ W and 8.0 x 10⁶ A of current was needed to produce this beam.

The voltage is calculated as

V= P / I

V=  (6.0 x 10¹³ ) / (8.0 x 10⁶)

V= 7.5 x 10⁶ Volts

Therefore, the voltage that produced the current is 7.5 x 10⁶ Volts.

Learn more about power.

brainly.com/question/2289248

#SPJ1

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Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
3 years ago
Please correct answers only, if you don’t know, ignore. Please answer this or am I correct?
Wewaii [24]

Answer:

C is correct.

Explanation:

5 0
3 years ago
Read 2 more answers
Which instrument is not used to measure air pressure
Virty [35]
It would be b. anemometer. it measures the speed of wind or any current of gas.
5 0
3 years ago
A 50 kg bicyclist, traveling at a speed of 12 m/s, applies the brakes, slowing her speed to 3 m/s.
Bezzdna [24]

a) Work done = Net Kinetic Energy

= 1/2 x 50 kg x ((12m/s)^2 - (3m/s)^2)

= 0.5 x 50 Kg x (144 -9)(m/s)^2

= 3375 Kg (m/s)^2

b) Force = mxa

a = 120 N/50 Kg = 2.4 m/s^2

Using newtons third law of motion, we get-

V^2 - U^2 = 2 x a x S

S= (12^2-3^2)m^2/s^2/(2 x 2.4 m/s^2)

= 28.125 m


3 0
3 years ago
A certain wire has a resistance of 110 Ω. What is the resistance of a second wire, made of the same material, that is 1/4 as lon
Svet_ta [14]

Answer:

New Resistance = 247.5 ohm

Explanation:

Resistance = resistivity * length / area

Since resistivity for the material is constant, resistance is directly proportional to (length/area).

This means that if (length/area) decreases or increases by any ratio, then resistance will increase or decrease by the same ratio.

So let's find the change in length/area

New length = 0.25 old length

New area = (1/9) old area                                 (This is because area equation contains a square of the diameter. if diameter decreases by 1/3, area decreases by (1/3)^2   )

So we now get length /area:

New length / new area = ( 0.25 old length) / (1/9 of old area)

New length / new area = 9*0.25 (old length / old area)

New length / new area = 2.25 (old length / old area)

To get the new resistance, we simply multiply it by the ratio we just found.

This equals:

110 * 2.25 = 247.5 ohm

4 0
3 years ago
Read 2 more answers
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