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kondor19780726 [428]

3 years ago

14

A scuba diver shines a flashlight from beneath the surface of water (n = 1.33 such that the light strikes the water-air boundary

with an angle of incidence of 37°. at what angle is the beam refracted?

1 answer:

Len [333]3 years ago

7 0

The beam is refracted at 53.17-degree angle (asin 0.800414 = 53.17-degree). This problem can be solved by using the Snell's Law which described the refracted beam angle which traveled through a different media, in this case, through water and air. The formula of Snell's Law is stated as n1 sin θ1= n2 sin θ2. In this formula n1 is the refraction index of medium 1, θ1 is the normal light angle in the medium 1, n2 is the refraction index of medium 2, and θ2 is the normal light angle in the medium 2<span>. </span>

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A net force of 50 N causes a mass to accelerate at a rate of 6.8 m/s2. Determine the mass.

Taya2010 [7]

Force=50N
Acceleration=6.8m/s^2

$\\ \sf\longmapsto F=ma$

$\\ \sf\longmapsto m=\dfrac{F}{a}$

$\\ \sf\longmapsto m=\dfrac{50}{6.8}$

$\\ \sf\longmapsto m=7.3kg$

5 0

3 years ago

Read 2 more answers

consider two stars, star a and star b. star a has a temperature of 4900 k , and star b has a temperature of 9900 k . how many ti

ValentinkaMS [17]

The Energy flux from Star B is 16 times of the energy flux from Star A.

We have Two stars - A and B with 4900 k and 9900 k surface temperatures.

We have to determine how many times larger is the energy flux from Star B compared to the energy flux from Star A.

<h3>State Stephen's Law?</h3>

Stephens law states that if E is the energy radiated away from the star in the form of electromagnetic radiation, T is the surface temperature of the star, and σ is a constant known as the Stephan-Boltzmann constant then-

$$\frac{Energy}{Area} = \sigma\times T^{4}$

Now -

Energy emitted per unit surface area of Star is called Energy flux. Let us denote it by E. Then -

$$E= \sigma\times T^{4}$

Now -

For Star A →

$T_{A}$ = 4900 K

For Star B →

$T_{B}$ = 9900 K

Therefore -

$$\frac{T_{B} }{T_{A} } =\frac{9900}{4900}$

$\frac{T_{B} }{T_{A} }=$ 2.02 = 2 (Approx.)

Now -

Assume that the energy flux of Star A is E(A) and that of Star B is E(B). Then -

$$\frac{E(B)}{E(A)} = \frac{\sigma\times T(B)^{4} }{\sigma \times T(A)^{2} }$

E(B) = E(A) x $(\frac{T(B)}{T(A)} )^{4}$

E(B) = E(A) x $2^{4}$

E(B) = 16 E(A)

Hence, the Energy flux from Star B is 16 times of the energy flux from Star A.

To learn more about Stars, visit the link below-

brainly.com/question/13451162

#SPJ4

4 0

2 years ago

In damped harmonic oscillation, the amplitude of oscillation becomes one third after 2 second. If A0 is initial amplitude of osc

Harman [31]

Answer:

$A=\frac{A_0}{\sqrt 3}$

Explanation:

Initial amplitude= $A_0$

We are given that

Amplitude after 2 s=A= $\frac{1}{3}A_0$

We have to find the amplitude after 1 s.

We know that amplitude at any time t

$A=A_0e^{-\alpha t}$

Using the formula

$\frac{A_0}{3}=A_0e^{-2\alpha}$

$\frac{1}{3}=e^{-2\alpha}$

$3=e^{2\alpha}$

$ln 3=2\alpha$

$\alpha =\frac{ln 3}{2}=ln\sqrt 3$

$e^{\alpha}=\sqrt 3}$

When t=1 s

$A=A_0e^{-\alpha}=\frac{A_0}{\sqrt 3}$

8 0

4 years ago

The difference between experimental technique and procedure

kaheart [24]

A procedure is all the steps used to do an experiment in order.
<span>the experiment is when you test your hypothesis and is designed to answer your question. </span>
<span>the procedure is all the steps of the experiment.</span>

4 0

4 years ago

calculate the density of a neutron star with a radius 1.05 x10^4 m, assuming the mass is distributed uniformly. Treat the neutro

Orlov [11]

To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

$\rho = \frac{m}{V}$

Where

m = Mass (Neutron at this case)

V = Volume

The mass of the neutron star is 1.4times to that of the mass of the sun

The volume of a sphere is determined by the equation

$V = \frac{4}{3}\pi R^3$

Replacing at the equation we have that

$\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}$

$\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}$

$\rho = 5.75*10^{17}kg/m^3$

Therefore the density of a neutron star is $5.75*10^{17}kg/m^3$

4 0

3 years ago