What is the average speed of a car that travels 15 km in 4000. seconds?

Which is true about total internal reflection?

Shkiper50 [21]

Answer:

It can occur only when light is incident on an interface where the index of refraction on the other side is less.

Explanation:

When the light passes from a denser medium, with refractive index n1, to another less dense medium, with refractive index n2, the incident light beam is refracted in such a way that it is not able to cross the surface between both media, the light beam is fully reflected and completely confining in the optically denser medium through which it propagates. For this phenomenon to occur, it is necessary that the angle of the incident light beam with respect to the normal be greater than or equal to the critical incidence angle θc. The critical angle can be calculated as :

$\theta_c=arcsin(\frac{n_2}{n_1})$

7 0

3 years ago

I need Help Fast ASAP<br> Also, SHOW YOUR WORK<br> WILL MARK BRAINLIEST!!!!<br> 100 points!!!

denis-greek [22]

Answer

7 is d ii is a iii is d

4 is x is 52 and y is 0 = 68

5 is x is 0 = 145 and y is 128

6 is a

1 is 10.20

2 is 151 and 5010

3 is < is 0 and I is y

9 is 3.0 and 2.0

hope this helps

6 0

3 years ago

A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it

soldi70 [24.7K]

Answer:

The position function is $s_{t}=2t^3+5t^2-5t+9$ .

Explanation:

Given that,

Acceleration $a =12t+10$

Initial velocity $v_{0} = -5\ cm/s$

Initial displacement $s_{0}=9\ cm$

We know that,

The acceleration is the rate of change of velocity of the particle.

$a = \dfrac{dv}{dt}$

The velocity is the rate of change of position of the particle

$v=\dfrac{dx}{dt}$

We need to calculate the the position

The acceleration is

$a_{t} = 12t+10$

$\dfrac{dv}{dt} = 12t+10$

$a_{t}=dv=(12t+10)dt$

On integration both side

$\int{dv}=\int{(12t+10)}dt$

$v_{t}=6t^2+10t+C$

At t = 0

$v_{0}=0+0+C$

$C=-5$

Now, On integration again both side

$v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt$

$s_{t}=2t^{3}+5t^2-5t+C$

At t = 0

$s_{0}=0+0+0+C$

$C=9$

$s_{t}=2t^3+5t^2-5t+9$

Hence, The position function is $s_{t}=2t^3+5t^2-5t+9$ .

7 0

3 years ago

What does a dielectric do to a capacitor?

Leto [7]

A dielectric fabric is used to split the conductive plates of a capacitor. This insulating cloth significantly determines the properties of a component. The dielectric constant of fabric determines the amount of strength that a capacitor can keep while voltage is applied.

A capacitor is an electrical component that attracts energy from a battery and stores the power. Internal, the terminals connect with metal plates separated by way of a non-engaging in substance. While activated, a capacitor quickly releases electricity in a tiny fraction of a 2d.

Capacitor, a tool for storing electric electricity, together with two conductors in near proximity and insulated from each different. A simple example of this sort of garage tool is the parallel-plate capacitor.

A capacitor is a tool that shops electric power in an electric-powered area. it is a passive digital thing with terminals. The effect of a capacitor is called capacitance.

Learn more about capacitors here brainly.com/question/13578522

#SPJ4

4 0

2 years ago

A cup of hot coffee initially at 95 degrees C cools to 80 degrees C in 5 min while sitting in a room of temperature 21 degrees C

Oksana_A [137]

Answer:

When the temperature of the coffee is 50 °C, the time will be 20.68 mins

Explanation:

Given;

The initial temperature of the coffee T₀ = 95 °C

The temperature of the room = 21°C

Let T be the temperature at time of cooling t in mins

According to Newton's law of cooling;

$\frac{dT}{dt} \alpha (T-21)\\\\\frac{dT}{dt} = k (T-21)\\\\\frac{dT}{T-21} = kdt\\\\\int\limits {\frac{dT}{T-21}} = \int\limits kdt\\\\Log(T-21) =kt + Logc \\\\Log (\frac{T-21}{c} ) = kt\\\\T -21 = ce^{kt}\\\\At \ t = 0, T = 95\\\\95-21 = ce^0\\\\74 = c\\\\New, equation: T -21 = 74e^{kt}\\\\Again; when \ t= 5\ min, T = 80\\\\80 -21 = 74e^{5k}\\\\59 = 74e^{5k}\\\\e^{5k} = \frac{59}{74}\\\\ 5k = ln(\frac{59}{74})\\\\5k = -0.2265\\\\k = -0.0453$

When the temperature is 50 °C, the time t in min is calculated as;

$T -21 = 74e^{-0.0453t}\\\\50 -21 = 74e^{-0.0453t}\\\\29 = 74e^{-0.0453t}\\\\\frac{29}{74} = e^{-0.0453t}\\\\0.39189 = e^{-0.0453t}\\\\ln(0.39189 ) = {-0.0453t}\\\\-0.93677 = {-0.0453t}\\\\t = \frac{-0.93677}{-0.0453}\\\\ t = 20.68 \ mins$

Therefore, when the temperature of the coffee is 50 °C, the time will be 20.68 mins

4 0

3 years ago