23692U = 14456Ba + 89 36Kr + 3n is a

A diver shines a flashlight upward from beneath the water (n = 1.33) at a 32.7° angle to the vertical. At what angle does the li

suter [353]

Answer:

45.93°

Explanation:

The angle of incidence is given as 32.7°

The refractive index of the water that is $n_1=1.33$

Refractive index of the air that is $n_2=1$ (because the refractive index of air is 1 )

We have to find the angle at which the light leave the water means angle of refraction

So according to snell's law $n_1sini=n_2sinr$

$1.33sin32.7=1\times sinr$

$sinr=0.7185$

r =45.93°

So the light leave the water at an angle of 45.93°

3 0

3 years ago

Which statement correctly describes the differences between positive and negative acceleration?

IceJOKER [234]

Answer: c) Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed.

Hey

On Earth, you can move or you can not move. if you are moving 50 mph that means relative to Earth (not the Andromeda galaxy). When you start moving (accelerating) you are now moving relative to Earth. when you start slowing down (decelerating [most scientists just say you have negative acceleration]) you are starting to match your velocity to Earth's velocity.

Hope it helped and made any sence at all.

Spiky bob your answerer.

3 0

3 years ago

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

kumpel [21]

Answer:

7800kg/m³

Explanation:

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

Given the density of iron = 7.8 g/cm3.

The SI units must be in kg/m³

7.8g = 7.8/1000 kg

7.8g = 0.0078kg

1cm³ = 0.000001m³

7.8g/cm³

= 0.0078/0.000001 kg/m³

= 7800kg/m³

Hence the density in SI unit is 7800kg/m³

4 0

3 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$