What is the velocity of the particle when its acceleration is zero?

Complete the ray diagram and label incident ray, refracted ray, angle of incidence, and angle of refraction

Novay_Z [31]

Answer:

Solution

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(a) The labelled diagram is shown.

(b) The refractive index of diamond is 2.42. Refractive index of diamond is the ratio of the speed of light in air to the speed of light in diamond.i.e.,

μ=

Speedoflightindiamond

Speedoflightinair

and, the ratio of these velocities is 2.42. i.e., This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air. In other words, the speed of light in diamond is

1/2.42

times the speed of light in vacuum.

Explanation:

a) Draw and label the diagram given :

(i) Incident ray

(ii) Refracted ray

(iii) Emergent ray

(iv) Angle of reflection

(v) Angle of deviation

(v) Angle of emergence

(b) The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?

4 0

3 years ago

A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'

Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = $\frac{1}{2}$ I ω^2 + $\frac{1}{2} mv^{2}$

where

I = $\frac{1}{2} mr^{2}$ and ω = $\frac{v}{r}$

thus,

KE = $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ ( $\frac{v}{r}$ )^2 + $\frac{1}{2} mv^{2}$

= $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ $\frac{v^{2} }{r^{2}}$ + $\frac{1}{2} mv^{2}$

= $\frac{1}{4} mv^{2}$ + $\frac{1}{2} mv^{2}$

= $\frac{3}{4} mv^{2}$

= $\frac{3}{4} (0.46) (1.1)^{2}$

= 0.42 J

8 0

3 years ago

Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol

klasskru [66]

The current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

J = (1.6 $\times$ 10^-19) / 0.001 (104 $\times$ 1200 + 1016 $\times$ 500)

J = 1012480 $\times$ 10^-16 A / m^2.

or

J = 1.01 $\times$ 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 $\times$ 10^-9 $\times$ 0.002 = 2.02 $\times$ 10^-12

So, the current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

6 0

3 years ago

Which of the following is an example of the conclusion phase of the scientific method?

Burka [1]

Answer:

a scientist examines the results and answers the lab question- last choice

5 0

3 years ago

A bullet with a mass of 5 gramsand speed of 560 m/sis fired horizontally atablock of wood with a mass of 2 kg. The block rests o

Anni [7]

Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.

So $mv=(m+M)v' \Leftrightarrow v'=\frac{mv}{m+M}$ .

Kinetic energy of system at first: $\frac{mv^2}{2}=784\;\textbf{J}$ . After: $\frac{(m+M)\frac{m^2v^2}{(m+M)^2} }{2}=\frac{m^2v^2}{2(m+M)}\approx 1,96\; \textbf{J}$ . The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).

Answer is A)

5 0

3 years ago