All categories

3 years ago

What happens to a light ray passing through the center of a lens?

Physics

2 answers:

lara31 [8.8K]3 years ago

7 0

What happens when a light ray passes through the focal point, and then reflects off of a convex mirror?

The reflected ray will run parallel to the center of curvature.

The reflected ray will run parallel to the principal axis.

The reflected ray will pass through the center of curvature.

The reflected ray will pass through a point between the center of curvature and the focal point.

The reflected ray will pass through a point between the focal point and the vertex.

Is it out of these choices?

Readme [11.4K]3 years ago

6 0

Theoretically, the light ray will not change direction. The center of a lens is called the optical center. If the ray of light hits the lens even just a little bit off center, the light ray is refracted. Refraction happens when the incident and exit angles of the ray are different. The optical center is guaranteed to not have any difference in angle, which allows the light ray to pass through directly without changing direction.

You might be interested in

The primary purpose of a switch in a circuit is to

Vanyuwa [196]

Answer:

See below

Explanation:

A switch regulates the transparency and closeness of an electrical circuit. This enables the current flow of the circuit to be regulated (not even possible to reach it and manually sever the wires). Switches are important components of any circuit that involve an input system or control

Hope this helps.

4 0

2 years ago

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km

ruslelena [56]

Answer:

x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

Angle which makes from ground = 58.9°

(a) X component of distance is given by $r_x=rcos\Theta =118\times cos58.9=118\times 0.5165=118=60.85m$

(b) Y component of distance is given by $r_Y=rcos\Theta =118\times sin58.9=118\times 0.8562=101.0316m$

These are the x and y component of position vector

6 0

3 years ago

5 points

olga2289 [7]

Answer:

d. 5 ohms

Explanation:

For resistors in parallel, the equivalent resistance is found with:

1/Req = ∑(1/R)

1/R = 1/15 + 1/15 + 1/15

1/R = 3/15

R = 15/3

R = 5

8 0

3 years ago

A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600

3241004551 [841]

Answer:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

$v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}$

Where:

$v_{o}$ - Initial speed of the rocket, in m/s.

$v_{ex}$ - Exhaust gas speed, in m/s.

$m_{o}$ - Initial total mass of the rocket, in kg.

$m$ - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

$m(t) = m_{o} + r\cdot t$

The initial total mass of the rocket is:

$m_{o} = 750\,kg$

The fuel consumption rate is:

$r = -\frac{600\,kg}{30\,s}$

$r = -20\,\frac{kg}{s}$

The function for the current total mass of the rocket is:

$m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t$

The speed function of the rocket is:

$v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}$

The speed of the rocket at given instants are:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

7 0

2 years ago

An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J.

AnnyKZ [126]

Answer:

a) $I=733.33\ A$

b) $d=52272.7273\ m$

c) $d'=51948.0519\ m$

Explanation:

Given:

voltage of the battery, $V=12\ V$

energy storage capacity of the battery, $E=2.3\times 10^7\ J$

speed of the car, $v=20\ m.s^{-1}$

power drawn by the car, $P=8.8\ kW$

Now the Current delivered to the motor:

we the relation between the power and electrical current,

$P=V.I$

$8800=12\times I$

$I=733.33\ A$

Distance travelled before battery is out of juice:

we first find the time before the battery runs out,

$t=\frac{E}{P}$

$t=\frac{2.3\times 10^7}{8800}$

$t=2613.636\ s$

Now the distance:

$d=v.t$

$d=20\times 2613.636$

$d=52272.7273\ m$

When the head light of 55 W power is kept on while moving then the power consumption of the car is:

$P'=P+55$

$P'=8800+55$

$P'=8855\ W$

Now the time of operation of the car:

$t'=\frac{E}{P'}$

$t'=\frac{2.3\times 10^7}{8855}$

$t'=2597.4026\ s$

Now the distance travelled:

$d'=v.t'$

$d'=20\times 2597.4025$

$d'=51948.0519\ m$

5 0

2 years ago