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natka813 [3]
3 years ago
11

Ok sooo I need help I don’t understand this lol

Physics
1 answer:
Burka [1]3 years ago
3 0

Answer:

I dont either but I'm sorry and hopefully you figured it out

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The mass of a moving object increases, but its speed stays the same. What happens to the kinetic energy of the object as a resul
Ronch [10]
We Know, K.E. = 1/2 × m × v²
From the expression, we can conclude that Kinetic energy is directly proportional to mass. So, as mass will increase, Kinetic energy will also increase.

In short, Your Correct answer would be Option B

Hope this helps!
8 0
3 years ago
Read 2 more answers
Closed clusters are groups of closely grouped stars that are located along the spiral disk of a galaxy.
Gre4nikov [31]
<span>False. Closed clusters are groups of closely grouped stars located in the halo of a galaxy, not along the spiral disk. Open clusters are groups of loosely associated stars located along the spiral disk of a younger galaxy with active star formation.</span>
5 0
3 years ago
Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses
Tom [10]

Answer:

a) U = 0.375 mJ

b) p_rad = 4.08 mPa

c) λ_med = 604 nm  ; f_med = 3.7 * 10^14 Hz

d) E_o  = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp  

Explanation:

Given:

λ_air : wavelength = 810 * 10^(-9) m

P: Power delivered  = 0.25 W

d : Diameter of circular spot = 0.00051 m

c : speed of light vacuum = 3 * 10^8 m/s

n_air : refraction Index of light in air = 1

n_med : refraction Index of light in medium = 1.34

ε_o : permittivity of free space = 8.85 * 10^-12 C / Vm

part a

The Energy delivered to retina per pulse given that laser pulses are 1.50 ms long:

U = P*t

U = (0.25 ) * (0.0015 )

U = 0.375 mJ

Answer : U = 0.375 mJ

part b

What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed?

                   

                                                p_rad = I / c

Where I : Intensity = P / A

                                                p_rad = P / A*c        

Where A : Area of circular spot = pi*d^2 / 4

                                               

                                                 p_rad = 4P / pi*d^2*c  

                              p_rad = 4(0.25) / pi*0.00051^2*(3.0 * 10^8)      

                                                 p_rad = 0.00408 Pa                          

Answer : p_rad = 4.08 mPa

part c

What are the wavelength and frequency of the laser light inside the vitreous humor of the eye?

                                    λ_med =  n_air*λ_air / n_med

                                     λ_med = (1) * (810 nm) / 1.34

                                           λ_med = 604 nm

                                               f_med = f_air

                                          f_med = c /  λ_air

                                 f_med = (3*10^8) / (810 * 10^-9)

                                          f_med = 3.7 * 10^14 Hz

Answer : λ_med = 604 nm  ; f_med = 3.7 * 10^14 Hz

d)

What is the electric and magnetic field amplitude in the laser beam?

                                        I = P / A

                                I  = 0.5*ε_o*c*E_o ^2

                                   I = 4P / pi*d^2

Hence,              E_o = ( 8 P /  ε_o*c*pi*d^2 ) ^ 0.5

E_o = ( 8 * 0.25 / (8.85*10^-12) * (3*10^8) * π * (0.00051)^2) ^ 0.5

                                 E_o  = 3.04 * 10^4 V / m

For maximum magnetic field strength:

                                      B_o = E_o / c

                         B_o = 3.04 * 10^4 / (3*10^8)

                         B _o = 1.013 *10^-4 Nm/Amp      

Answer: E_o  = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp      

5 0
3 years ago
A 40.0 turn coil of wire of radius 3.0 cm is placed between the poles of an electromagnet. The field increases from 0 to 0.75 T
nadezda [96]

Answer:

Induced emf, V=3.76\times 10^{-4}\ V

Explanation:

We have,

Number of turns in the coil, N = 40

Radius of coil, r = 3 cm = 0.03 m

The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s.

It is required to find the magnitude of the induced emf in the coil if the field is perpendicular to the plane of the coil. The induced emf is given by :

\epsilon=\dfrac{d\phi}{dt}

\phi=NBA\cos\theta, is magnetic flux

\epsilon=NA\dfrac{dB}{dt}\\\\\epsilon=\dfrac{40\times \pi (0.03)^2\times (0.75-0)}{225}\\\\\epsilon=3.76\times 10^{-4}\ V

So, the magnitude of induced emf is 3.76\times 10^{-4}\ V.

7 0
3 years ago
Two blocks with different temperatures had entropies of 10 J/K and 30 J/K before they were brought in contact. What can you say
True [87]

Answer:

a. Ssystem  > 40 J/K

Explanation:

Given that

The entropy of first block = 10 J/K

The entropy of second block = 30 J/K

When two bodies come into contact with each other, the entropy of the combined system will increase and the entropy sum remains unchanged: According to the Second law of thermodynamics.The entropy of the system will be greater than 40  J/K.

Therefore the answer is a.

Ssystem  > 40 J/K

6 0
3 years ago
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