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EleoNora [17]
2 years ago
9

Do electromagnetic waves require matter to travel?

Physics
1 answer:
VMariaS [17]2 years ago
6 0

Answer:

No.

Explanation:

Electromagnetic waves do not require a medium of matter to move through, electromagnetic waves are used in things like your cell phone and telecommunications.

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Calculate the most probable speed of an ozone molecule in the stratosphere
Marysya12 [62]

Answer:

v_{mp}=305.83 m/s

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

v_{mp}= \sqrt{\frac{2RT}{M} }

plugging the values we get

v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }

v_{mp}=305.83 m/s

7 0
3 years ago
When a certain isotope, such as U-238, is hit by a neutron, it will always split into the same smaller nuclei.
nadezda [96]

Answer: falss

Explanation:

4 0
3 years ago
Read 2 more answers
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
James weighs 67 pounds on Earth. He has a mass of 30 kilograms. Which of the following is true?
morpeh [17]

Answer:

I think its B. James would have a smaller mass on the Moon than he does on Earth.

sorry if i did it wrong

4 0
3 years ago
Read 2 more answers
A sealed box contains a monatomic ideal gas. The number of gas atoms per unit volume is 5.00 * 1020 atoms>cm3, and the averag
Veronika [31]

Answer:

Pressure,P=6×10^3Pa

Explanation:

The gas has an ideal gas behaviour and ideal gas equation

PV=NKT

T= V/N p/K ...eq1

Average transitional kinetic energy Ktr=1.8×10-23J

Ktr=3/2KT

T=2/3Ktr/K....eq2

Equating eq1 and 2

V/N p/K = 2/3Ktr/K

Cancelling K on both sides

P= 2/3N/V( Ktr)

Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3

P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23

P= 6 ×10^3Pa

6 0
3 years ago
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