PLZZZ HELP

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible

Leto [7]

Answer:

a) 298.5 nm , 522.4 nm and b) radiation frequency does not change

Explanation:

When electromagnetic radiation reaches a medium with a different index of refraction, the medium vibrates the molecules, as if it were a resonance process, whereby the medium vibrates at the same frequency as the incident light.

On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio

λₙ = λ₀ / n

λₙ = 400 nm in the void

λₙ = 400 / 1.34

λₙ= 298.5 nm

λ₀ = 700 nm

λₙ = 700 / 1.34

λₙ = 522.4 nm

The radiation frequency does not change

4 0

3 years ago

Help it’s multiple choice 11 through 15 please!

riadik2000 [5.3K]

1. • Here, force of gravity on the block = 20 N.

• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]

• The coefficient

$u_{k} = 0.4$

• Force of friction =

$u_{k} \times \: normal \: \: \: force \\ = 0.4 \times 20N \\ = 8N$

• Hence, the force of sliding friction between the block and the ground is 8 N.

• So, it is option c. 8 N

2. The answer is option d. continue in the same direction with no change in speed.

We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.

3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.

Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.

So, 1 N = 1 Kg × 1 m/s^2.

4. It is d. not zero.

Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.

5. Force = 2 N

Acceleration of the object A = 2 m/s^2.

Acceleration of the object B = 1 m/s^2.

Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg

And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg

So, the mass of object B is greater than that of object A.

Hence, the answer is option c. Object B has more mass.

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0

2 years ago

Which of the following terms corresponds to #2 on the image?

Mila [183]

Answer:

Trough

Explanation:

cuz physics you see

7 0

3 years ago

A skateboarder shoots off a ramp with a velocity of 7.1 m/s, directed at an angle of 61° above the horizontal. The end of the ra

dsp73

Answer:

Highest point reached = 3.37 m

Explanation:

Initial velocity, = 7.1 m/s

Initial vertical velocity = 7.1 sin 61 = 6.21 m/s

Consider the vertical motion of skateboarder,

We have equation of motion, v² = u² + 2as

Initial velocity, u = 6.21 m/s

Acceleration, a = -9.81 m/s²

Final velocity, v = 0 m/s

Substituting

v² = u² + 2as

0² = 6.21² + 2 x -9.81 x s

s = 1.97 m

So from ramp the position it goes up by 1.97 m

Highest point reached = 1.97 + 1.4 = 3.37 m

6 0

3 years ago

A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2

stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

$s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2$

For the bottom of window (position B)

$s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2$

$\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673$

We also have

$t_B-t_A=0.27$

Solving

$t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s$

So after 1.11 seconds ball reaches at top of window,

We have equation of motion v = u + at

$v_A=0+9.81\times 1.11=10.89m/s$

Speed at which the ball passes the window’s top = 10.89 m/s

7 0

3 years ago