A rock is dropped from a height of 3.4 m. How much time does it take to hit

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

All waves have wavelength, frequency, rest point, and speed. <br> True or False?

nexus9112 [7]

Answer:

it is true

Explanation:

4 0

3 years ago

Read 2 more answers

How can people reconcile technology with environmental preservation?

MatroZZZ [7]

1. Developing renewable energy technology

- Efficient energy storage and smarter grids .
- renewable and rechargeable batteries and fuel cell

2. Saving endangered wildlife

-smart collars for endangered species and reducing human - animal conflict
-Gene sequencing for detecting and researching on deadly animal diseases.

3. Adopting a smarter lifestyle

- smart homes that promote energy saving and green - living .
- electric cars which are three times more conventional vehicles .

Hope this helps :)

5 0

3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago

What is static friction?

givi [52]

Answer:

Plzzzzzzzzzzzzzzzz brainliest

Explanation:

In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object. ... The frictional force itself is directed oppositely to the motion of the object.

7 0

2 years ago