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3 years ago

11

What is an incident angle

2 answers:

hodyreva [135]3 years ago

8 0

Answer:

The angle between a ray incident on a surface and the line perpendicular to the surface at the point of incidence.

Explanation

Hunter-Best [27]3 years ago

5 0

Answer:In geometric optics, the angle of incidence is the angle between a ray incident on a surface and the line perpendicular to the surface at the point of incidence, called the normal. The ray can be formed by any wave: optical, acoustic, microwave, X-ray and so on. In the figure below, the line representing a ray makes an angle θ with the normal. The angle of incidence at which light is first totally internally reflected is known as the critical angle. The angle of reflection and angle of refraction are other angles related to beams.

Explanation:tik tok: Uh.amy07

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A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic

olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma

$a = \frac{F}{m}$ ...... (1)

acceleration (a) $= \frac{dv}{dt}$ ......(2)

substituting (2) into (1)

Hence, F $= \frac{mdv}{dt}$

$\frac{dv}{dt} = \frac{F}{m}$

$dv = \frac{F}{m} dt$

$dv = \frac{1}{m}Fdt$

Integrating both sides

$\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt$

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

$v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt$ ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

$v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt$

$v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}$

$v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|$

$v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |$

$v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |$

$v = \frac{1}{5} | 250 - 50 + 15 |$

$v = \frac{215}{5}$

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

6 0

3 years ago

What is dispersion of light?

algol13

Answer:

$\huge \bold \blue{ \underline{ answer}}$

The splitting up of light into its constituent colours while passing from one medium to the other is called dispersion.

3 0

2 years ago

Read 2 more answers

Is demand for tea elastic or inelastic

JulsSmile [24]

Tea elastic, i think that is right? if not please tell me??!!!!

5 0

4 years ago

(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model i

Anni [7]

Answer:

(a) $L=2.6742\times 10^{40}\, kg.m^2.s^{-1}$

(b) $L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}$

Explanation:

<u>For Earth we have:</u>

mass of earth, $m_E=5.97\times 10^{24}\, kg$

radius of earth, $R_E=6.38\times 10^6m$

orbital radius, $r=1.5 \times 10^{11}m$

period of rotation, $t_{rot}=24h=86400\, s$

period of revolution, $t_{rev}= 1 yr=3.156\times 10^7 s$

(a)

Angular momentum, $L=?$

∵ $L=I.\omega$ ...............................(1)

For a particle of mass m moving in a circular path at a distance r from the axis,

$I=m.r^2$

& $v=r.\omega$

Putting respecstive values in eq. (1)

$L=m_E\times r^{2}\times \omega$

$L=5.97\times 10^{24}\times (1.5 \times 10^{11})^2\times \frac{2\pi}{3.156\times 10^7}$

$L=2.6742\times 10^{40}\, kg.m^2.s^{-1}$

To model earth as a particle is reasonable because the distance between the sun and the earth is very large s compared to the radius if the earth.

(b)

For a uniform sphere of mass M and radius R and an axis through its center, $I=\frac{2}{5}M.R^2$

using eq. (1)

$L_s=(\frac{2}{5}m_E.R_E^2) \omega$

$L_s=\frac{2}{5} \times 5.97\times 10^{24} \times 6.38\times 10^6\times \frac{2\pi}{86400}$

$L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}$

6 0

3 years ago

Scientists described light as a wave because the results of many experiments with light demonstrated that light behaves as a wav

seropon [69]

B and C are both correct.

5 0

3 years ago