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kirill115 [55]
3 years ago
12

Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of

electromagnetic waves are these? (c) How much energy (in electron volts) is carried by one quantum of this radiation?

Physics
1 answer:
laiz [17]3 years ago
8 0

Answer:

a) the frequency is 3.191x10^11

b)the radiation is a microwave radiation

c) the energy of one quantum of this radiation is 1.32x10^-3ev

Explanation:

Detailed explanation and calculation is shown in the image below

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Dmitry_Shevchenko [17]

Answer:

Doubling the voltage in this arrangement both doubles the voltage drop across the resistor and the current through it. The bulb will be much brighter.

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2 years ago
The equation most often associated with Newton's Second Law of Motion is<br> Answer here
zloy xaker [14]

Answer: F=ma

Explanation:

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2 years ago
Read 2 more answers
1. A 0.40 kg ball is launched at a speed of 16 m/s from a 22 m cliff.
Masja [62]

Answer:

51.2 J, 86.2 J, 137.4 J

Explanation:

The kinetic energy of the ball is given by:

K=\frac{1}{2}mv^2

where

m = 0.40 kg is its mass

v = 16 m/s is its speed

Substituting,

K=\frac{1}{2}(0.40)(16)^2=51.2  J

The potential energy of the ball is given by

U=mgh

where

m = 0.40 kg

g=9.8 m/s^2 is the acceleration of gravity

h = 22 m is the heigth of the cliff

Substituting,

U=(0.40)(9.8)(22)=86.2 J

Finally, the total mechanical energy is the sum of the kinetic energy and the potential energy:

E=K+U=51.2 + 86.2=137.4 J

4 0
2 years ago
A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i
Effectus [21]

Answer:

s =  1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore  

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}  

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}  

• s = \sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}

s = \sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}

s =  1.7 m

7 0
3 years ago
A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
maks197457 [2]

Answer:

5.6\times 10^{-8}\ Ohm.m

Explanation:

Resistivity is given by \rho=\frac {AR}{L} where A is cross-sectional area, R is resistance, L is the length and \rho is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

8 0
3 years ago
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