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makvit [3.9K]
3 years ago
9

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T

hey are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of 1.52\times 10^{11}~\text{V/m}1.52×10 ​11 ​​ V/m for a time of 1.00 ns. What energy does it deliver on a 1.00~\mathrm{mm^2}1.00 mm ​2 ​​ area?
Physics
1 answer:
alex41 [277]3 years ago
4 0

Answer:

30643 J

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

t = Time taken = 1 ns

c = Speed of light = 3\times 10^8\ m/s

E_0 = Maximum electric field strength = 1.52\times 10^{11}\ V/m

A = Area = 1\ mm^2

Magnitude of magnetic field is given by

B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T

Intensity is given by

I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2

Power, intensity and time have the relation

E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J

The energy it delivers is 30643 J

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MissTica

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Earthen materials were worn away during erosion, a geological process in which they are moved by water or wind. Weathering, a related process that does not involve movement, dissolves and breaks down rock.

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5 0
1 year ago
A coin is dropped off of a building landing on its side. It hits with a pressure of 400/2 It hits with a force of 0.1 Calculate
Arte-miy333 [17]

Complete Question:

A coin is dropped off of a building landing on its side. It hits with a pressure of 400 N/m². It hits with a force of 0.1N. Calculate the area of the coin?

Answer:

Area = 0.00025 m²

Explanation:

Given the following data;

Pressure = 400N/m²

Force = 0.1N

To find the area of the coin;

Pressure = Force/area

Area = Force/pressure

Substituting into the equation, we have;

Area = 0.1/400

Area = 0.00025 m²

5 0
3 years ago
An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide
Dmitry_Shevchenko [17]

Answer:

(a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2\ cm

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

v=\sqrt{\dfrac{2eV}{m}}

Put the value into the formula

v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}

v=15689290.81\ m/s

v=1.56\times10^{7}\ m/s

(b). We need to calculate the radius of electron

Using formula of centripetal force

\dfrac{mv^2}{r}=qvB

r=\dfrac{mv}{qB}

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}

r=0.042\ m

r=4.2\ cm

Hence, (a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2 cm

8 0
3 years ago
Over a span of 6.0 seconds, a car changes its speed from 89 km/h to 37 km/h. What is its average acceleration in meters per seco
faltersainse [42]

Answer: Average acceleration is -2.407 \frac{m}{s^{2}}

 

Explanation:

Average acceleration is equal to change in velocity divided by time taken

therefore, a=\frac{Vf-Vi}{t}

=> a=\frac{37-89}{6}\times \frac{5}{18}\frac{m}{s^{2}} = -2.407 \frac{m}{s^{2}}

6 0
4 years ago
What net force is needed to increase the velocity of a 5 kg object by 3 m/s over the course of 2.5 seconds?
mixer [17]

Hi there!

Recall Newton's Second Law:

\large\boxed{\Sigma F = ma}

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We must begin by solving for the acceleration using the following:

a = Δv/t

In this instance:

Δv = 3 m/s

t = 2.5 sec

a = 3/2.5 = 1.2 m/s²

Now, plug this value along with the mass into the equation for net force:

\Sigma F = 5(1.2) = \boxed{6 N}}

6 0
3 years ago
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