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2 years ago

If a radio wave has a period of 1 μs what is the wave's period in seconds

Physics

1 answer:

Nezavi [6.7K]2 years ago

7 0

Answer:

$10^{-6} s
$

Period of wave is defined by the time taken to complete one cycle of the wave i.e. from one crest to one trough.

Since, the period of the wave is already given in microseconds, we just have to convert into seconds.

we know

$1\mu s=10^{-6} s
$

Therefore, the period of wave would be $10^{-6} s
$ .

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An argon ion laser puts out 5.0 W of continuous power at a wavelength of 532 nm. The diameter of the laser beam is 5.5 mm. If th

stepan [7]

Answer:

Number of photons travel through pin hole= $6.4*10^{17}$

Explanation:

First we will calculate the energy of single photon using below formula:

$E=\frac{h*c}{λ}$

Where :

h is plank's constant with value $6.626*10^{-34} J.s$

c is the speed of light whch is $3*10^{8}$

λ is the wave length = 532nm

$E=\frac{6.6268*10^{-34}* 3*10^{8}}{532nm}$

E= $3.73*10^{-19}$ J

Number of photons emitted per second:

$\frac{5J/s}{1 photon/3.73*10^{-19} }$

Number of photons emitted per second= $1.34*10^{19} photons/s$

$\frac{A-hole}{A-beam}$ = $\frac{\frac{pie*d-hole^2}{4}}{\frac{pie*d-beam^2}{4} }$

Where:

A-hole is area of hole

A-beam is area of beam

d-hole is diameter of hole

d-beam is diameter if beam

$\frac{A-hole}{A-beam}$ = $\frac{d-hole^2}{d-beam^2}$

$\frac{Ahole}{A-beam}$ = $\frac{1.22^2}{5.5^2}$

$\frac{A-hole}{A-beam}$ = $\frac{144}{3025}$

Number of photons travel through pin hole= $1.34*10^{19} *\frac{144}{3025}$

Number of photons travel through pin hole= $6.4*10^{17}$

7 0

2 years ago

Calculate the amount of energy produced in a nuclear reaction in which the mass defect is 0.187456 amu.

luda_lava [24]

For nuclear reactions, we determine the energy dissipated from the process from the Theory of relativity wherein energy is equal to the mass defect times the speed of light. We calculate as follows:

E = mc^2 = 0.187456 (3x10^8)^2 = 1.687x10^16 J

Hope this answers the question.

8 0

2 years ago

Read 2 more answers

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th

mote1985 [20]

Answer:

U = 1794.005 × 10⁶ J

Explanation:

Data provided;

Capacitance of the original capacitor, C = 1.27 F

Potential difference applied to the original capacitor, V = 59.9 kV

= 59.9 × 10³ V

Now,

The Potential energy (U) for the capacitor is calculated as:

Potential energy of the original capacitor, U = $\frac{\textup{1}}{\textup{2}}$ × C × V²

on substituting the respective values, we get

U = $\frac{\textup{1}}{\textup{2}}$ × 1.27 × ( 59.9 × 10³ )²

U = 1794.005 × 10⁶ J

7 0

2 years ago

A child blows a leaf straight up in the air from rest. The leaf accelerates at 1.0\,\dfrac{\text m}{\text s^2}1.0 s 2 m1, point,

Neko [114]

Answer:

Time taken by the leaf to displace by 1.0 m distance is

$t = \sqrt2$ seconds

Explanation:

As we know that initial velocity of the leaf is given as

$v_i = 0$

now the acceleration upwards for the leaf is

$a = 1 m/s^2$

The displacement of leaf in upward direction is

d = 1 m

so now we have

$d = v_i t + \frac{1}{2}at^2$

$1 = 0 + \frac{1}{2}(1) t^2$

$t = \sqrt2$ seconds

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2 years ago

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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.