Jail

Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When

kupik [55]

Answer:

therefore critical angle c= 69.79°

Explanation:

Canola oil is less dense than water, so it floats over water.

Given $n_{canola}= 1.47$

which is higher than that of water

refractive index of water $n_{water}=1.33$

to calculate critical angle of light going from the oil into water

we know that

$sinc= \frac{n_{water}}{n_{canola}}$

now putting values we get

$sinc= \frac{1.33}{1.47}$

c= $sin^{-1}(\frac{1.33}{1.47} )$

c=69.79°

therefore critical angle c= 69.79°

8 0

3 years ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago

Please someone help!!!

antoniya [11.8K]

I know it’s the Coulomb’s law and that I’m pretty sure the answer would be C.Inverse Square.

4 0

3 years ago

Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k

NikAS [45]

Answer:

$Fr=12544N$

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation $Fr=usmg$ , where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

$Fr=0.8*1600kg*9.8\frac{m}{s^{2}}$

$Fr=12544N$

5 0

3 years ago

give an example of motion in which displacement of the particle is zero but acceleration is not zero in journey?

BartSMP [9]

Motion of a ball thrown by a person upwards and caught after some time is an example of motion in which displacement of the particle is zero but acceleration is not zero in journey.

The displacement of the ball is zero because the starting and end point of the motion are same, i.e, the person's hands.During its motion, the acceleration of ball is constant and non zero called as acceleration due to gravity, g= -9.8 m/s². The velocity of ball is continuously changing. It first decreases during the upward motion of the ball and then increases during the downward journey.The acceleration remains constant and non zero all the time.

4 0

3 years ago