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DerKrebs [107]
2 years ago
12

Even though this region is gaseous, it is regarded as the Sun’s surface because at this point, light from the Sun is finally abl

e to ________________________ .
a
get warm
b
turn into a liquid
c
escape into the solar system
d
to turn into a solid
Physics
1 answer:
galina1969 [7]2 years ago
7 0
C- escape into the solar system, because the sun is neither a solid nor a liquid, and the sun already creates its warm temperature from many reactions. the light from the sun can scatter throughout the universe, eventually getting to earth. do, have you ever heard of a wave of light being referred to as a solid or a liquid?
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When a piece of copper is taken to the moon , a change will be observed in it's ?​
SpyIntel [72]
Weight. Because there is less gravity on the moon.
3 0
3 years ago
A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with
Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D =  1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7}  m^{2}\\

Intensity

I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }

Amplitude of Electric Field

E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C

Amplitude of Magnetic Field

B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{  3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T

5 0
4 years ago
Which change of state do particles in a material become farther apart?
crimeas [40]
Particles become farther apart from each other in the change of state between liquid and gas.
Hope that helped =)
6 0
3 years ago
An electric dipole consists of charges +2e and -2e separated by 0.82 nm. It is in an electric field of strength 3.2 x 10^6 N/C.
Zepler [3.9K]

Explanation:

It is given that,

An electric dipole consists of charges +2e and -2e separated by 0.82 nm

Charge, q=2e=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C

Distance between charges, d=0.82\ nm=0.82\times 10^{-9}\ m

Electric field strength, E=3.2\times 10^6\ N/C

(a) The magnitude of the torque on the dipole is given by :

\tau=p\times E\ sin\theta

When dipole moment is parallel to the electric field, \theta=0

\tau=p\times E\ sin(0)

\tau=0

(b) When the dipole is perpendicular to the electric field, \theta=90

\tau=pE\ sin(90)

\tau=qdE (Since, p = q × d)

\tau=3.2\times 10^{-19}\times 0.82\times 10^{-9}\times 3.2\times 10^6

\tau=8.39\times 10^{-22}\ N.m

(c) When the dipole moment is anti parallel to the electric field, \theta=180

\tau=pE\ sin(180)

Since, sin\ 180=0

\tau=0

Hence, this is the required solution.

8 0
4 years ago
The roller-coaster car shown in fig. 6-41 (h1 = 45 m, h2 = 16 m, h3 = 26 m), is dragged up to point 1 where it is released from
kirill [66]

There are many ways to solve this but I prefer to use the energy method. Calculate the potential energy using the point then from Potential Energy convert to Kinetic Energy at each points.

PE = KE

From the given points (h1 = 45, h2 = 16, h<span>3  </span>= 26)

Let’s use the formula: 

v2= sqrt[2*Gravity*h1]  where the gravity is equal to 9.81m/s2

v3= sqrt[2*Gravity*(h1 - h3 )] where the gravity is equal to 9.81m/s2

v4= sqrt[2*Gravity*(h1 – h2)] where the gravity is equal to 9.81m/s2

Solve for v2

v2= sqrt[2*Gravity*h1]      

    = √2*9.81m/s2*45m

v2= 29.71m/s

v3= sqrt[2*Gravity*(h1 - h3 )   

    =√2*9.81m/s2*(45-26)

    =√2*9.81m/s2*19 

v3=19.31m/s

v4= sqrt[2*Gravity*(h1 – h2)]        

    =√2*9.81m/s2*(45-16)

    =√2*9.81m/s2*(29)

v4=23.85m/s

7 0
3 years ago
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