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stepladder [879]

3 years ago

5

As the change in speed at the boundary of two materials is greater what happened to the angle of refraction (please help )

1 answer:

Studentka2010 [4]3 years ago

7 0

The answer is B.
Hope this helps.

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If a football player collides with a goal post, what forces are at work?

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When a footballer collides with the goal post, the forces at work are the action and reaction forces. The player will exert an action force on the goal post, and then a reaction force from the goal post will stop the player. The reaction force call will cause pain and even injury to the player.

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A non-_____ rock has interlocking grains with no specific pattern.

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A non <span>foliated </span>rock has interlocking grains with no specific pattern.

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A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$

There is no kinetic energy in the initial state, nor potential energy in the end,

$mgh+0=0+KE_f$

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

$mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)$

The inertia of the bodies is given by the equation,

$I=\frac{m(R_1^2+R^2_2)}{2}$

$I=\frac{2(0.2^2+0.1^2)}{2}$

$I=0.05Kgm^2$

On the other hand the angular velocity is given by

$\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s$

Replacing these values in the equation,

$(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2$

Solving for h,

$h=2.86m$

5 0

3 years ago

How could you record the number 4000 and report 2 significant figures?

yawa3891 [41]

Explanation:

Write in scientific notation.

4000 = 4.0×10³

5 0

3 years ago

(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel

sammy [17]

Answer:

Part a)

$R_1 = 0.072 m$

Part b)

$R_2 = 0.036 m$

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

$R = \frac{mv}{qB}$

now for single ionized we have

$R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}$

$R_1 = 0.072 m$

Part b)

Similarly for doubly ionized ion we will have the same equation

$R = \frac{mv}{qB}$

$R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}$

$R_2 = 0.036 m$

Part c)

The distance between the two particles are half of the loop will be given as

$d = 2(R_1 - R_2)$

$d = 2(0.072 - 0.036)$

$d = 0.072 m$

6 0

3 years ago