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Mariulka [41]
2 years ago
9

As light shines from air to another medium, i = 26.0 º. The light bends toward the normal and refracts at 32.0 º. What is the in

dex of refraction? A. 1.06 B. 0.944 C. 0.827 D. 1.21
Physics
1 answer:
finlep [7]2 years ago
4 0

Explanation:

It is given that,

Angle of incidence from air to another medium, i = 26°

The angle of reflection, r = 32°

We need to find the refractive index of the medium. The ratio of sine of angle of incidence to the sine of angle of reflection is called refractive index. It can be given by :

n=\dfrac{\sin i}{\sin r}\\\\n=\dfrac{\sin (26)}{\sin (32)}\\\\n=0.82

So, the index of refraction is 0.82. Hence, the correct option is C.

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A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?
guajiro [1.7K]

The correct answer to the question is-  2.2\ \mu C

CALCULATION:

As per the question, the electric field generated by the source charge is 1236  N/C at a distance of 4 m.

Hence , electric field  E =  1236 N/C.

The distance of the point R = 4m

We are asked to calculate the charge possessed by the source.

The electric field produced by a source charge of Q at a distance R is calculated as -

                    Electric field E = \frac{1}{4\pi \epsilon_{0}}\frac{Q}{R^2}

Here, \epsilon_{0} is called the absolute permittivity of the free space.

Hence, the charge of source is calculated as -

                                         Q = E\times 4\pi \epsilon_{0}\times R^2

                                            = 1236\times \frac{1}{9\times 10^9}\times (4)^2\ Coulomb

                                            = 2197.33\times 10^{-9}\ C

                                             = 2.19733\times 10^{-6}\ C

                                             = 2.2\ \mu C

Hence, the charge of source is 2.2\ \mu C

3 0
2 years ago
Read 2 more answers
on the surface of planet x a body with a mass of 10 kilograms weighs 40 newtons. The magnitude of the acceleration due to gravit
tamaranim1 [39]
Based on the Newton's second law of motion, the value of the net force acting on the object is equal to the product of the mass and the acceleration due to gravity. If we let a be the acceleration due to gravity, the equation that would allow us to calculate it's value is,
      W = m x a
where W is weight, m is mass, and a is acceleration. Substituting the known values,
    40 kg m/s² = (10 kg) x a
Calculating for the value of a from the equation will give us an answer equal to 4. 
ANSWER: 4 m/s². 

6 0
2 years ago
Read 2 more answers
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Juliette [100K]

Answer:

24.348mm

Explanation:

NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards

K = d / €^n

Note : d represents the greek alphabet epsilion.

K = 345 / 0.02⁰.²² = 816mPa

The true strain based upon the stress of 414mPa =

€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576

However the true relationship between true strain and length is given by

€ = ln(Li/Lo)

Making Li the subject of formula by rearranging,

Li = Lo.e^€

Li = 520e⁰.⁰⁴⁵⁷⁶

Li = 544.348mm

The amount of elongation can be calculated from

Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.

8 0
2 years ago
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
kipiarov [429]

Answer:

I=68.31\times 10^{-6}\ A

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

By putting the values

I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A

I=68.31\times 10^{-6}\ A

7 0
2 years ago
Read 2 more answers
Students use a stretched elastic band to launch carts of known mass horizontally on a track. The elastic bands exert a force F,
rodikova [14]

Answer:

the correct answer is  E

A graph of the cart's maximum speed squared as a function of x^3

Explanation:

For this exercise let's use Newton's second law

        F = m a

force has the form

        F = k x²

and acceleration is related to velocity

        a = dv / dt

Let's use the chain rule or L'Hospital

        a = dv /dx   dx/dt

        a = dv /dx    v

let's substitute

     k x² = m v dv / dx

     k /m x² dx = v dv

we integrate

     k /m    x³ /3 = v² / 2

     v² = (2k /3m)   x³

This is the expression for the variation of the speed as a function of the position, to make a linear graph realism the changes of variable

      y = v²

     x´ = x³

       y = (2k/3m)  x´

if we graph y vs x 'we have a linear graph whose slope is

      m = 2k / 3m

By reviewing the different answers, the correct answer is  E

4 0
2 years ago
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