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Salsk061 [2.6K]
3 years ago
9

If dot C moves to position 5 (60 meters), what is its average velocity during these 5 seconds? A) 10 m/s B) 11 m/s C) 12 m/s D)

13 m/s
Physics
2 answers:
BartSMP [9]3 years ago
8 0

Answer:

C) 12 m/s

Explanation:

As we know that average velocity is given as ratio of displacement and time

so here we can say

v_{avg} = \frac{displacement}{time}

now we have

displacement = 60 meter

time = 5 seconds

now as per above formula we have

v_{avg} = \frac{60}{5}

v_{avg} = 12 m/s

so the average velocity must be 12 m/s

nlexa [21]3 years ago
6 0
C) 12 m/s because 60 m ÷ by 5sec equals 12m/s
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What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh
Goshia [24]

Answer:

a) R_s = 0.092

b) R_p = 0.085

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

R_s = 0.092

b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

R_p = 0.085

3 0
3 years ago
1. Plot the following graphs:
VLD [36.1K]

Answer:

(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram

(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

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Explanation:

8 0
3 years ago
What are some ways you can vary your tone of voice to help you communicate effectively with others? A. Speaking aggressively
Leno4ka [110]
The answer is D, talking more loudly or quietly.
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An expressive voice can bring totally different meaning to what you are saying. If you speak in a monotone voice, you are implying that you do not care about what you are saying or what the other person is talking about. If you speak expressively with some parts quieter or louder to make a point, it lets the listener know that you care about what you are saying and you care about what they have to say too. You can also emphasize a point by speaking louder or indicate that something is not as important by speaking softer.</span>

7 0
3 years ago
Read 2 more answers
Select the properties that apply to superconductors.
makkiz [27]
made from pure metals . . . no;
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conduct electricity with zero resistance . . . yes;
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just like any other conductor, the magnetic field depends on the current that's flowing in the conductor.

no loss of energy in the transfer of electricity . . .
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7 0
3 years ago
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A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit
anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}

            = 26.6 e V

h c = 6.626 ×  10⁻³⁴ ×  3  × 10⁸  = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁  = 23.2 + 13.6

       = 36.8 e V

energy of the original photon is 36.8 eV

3 0
3 years ago
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