A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cm
B. 60 cm
C. 75 cm
D. 90 cm

Answers

Answer 1

Answer: B. 60 cm

All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.

Related Questions

Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?
(PLEASE HELP ITS DUE SOON ILL MARK BRAINLIEST AND 5 STARS & PLEASE SHOW WORK!!)(And the answer is not 44 I already tried that and it doesn’t start with 4 either)
A proton and an alpha particle (helium nucleus consisting of two protons and two neutrons) are accelerated from rest across the same potential difference. Assume the proton mass and the neutron mass are roughly the same and neglect any relativistic effect. Compared to the final speed of the proton, the final speed of the alpha particle is?1. less by a factor of 22. less by a factor of √ 23. less by a factor of 44. greater by a factor of 25. the same
The pedals on a bicycle give a mechanical advantage by allowing you to turn the pedals a __________ distance to turn the __________ circumstance of the wheels.
. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.You should obtain e/m = 2V/(B^2)(r^2)3. The magnetic field on the axis of a circular current loop a distance z away is given byB = mu I R^2 / 2(R^2 + z^2)^ (3/2)where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/Rwhere B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done by the force

Answers

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

$\Rightarrow a=(F)/(m)$

Hence acceleration of the block is given by:

$\Rightarrow a=(25)/(2)=12.5m/s^2$

Displacement of the object is given by:

$\Rightarrow S=ut+(1)/(2)at^2$

Substitute the values

$\Rightarrow S=0*3+(1)/(2)(12.5)3^2$

$\Rightarrow S=56.25 m$

Now work done is given by:

W=F.S

W = 25×56.25

W =1406.25 J

Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.

Answers

Answer:

The period of the pendulum is $T = 1.68 \ sec$

Explanation:

The diagram illustrating this setup is shown on the first uploaded image

From the question we are told that

The length of the rod is $L = 80 \ cm$

The diameter of the ring is $d = 10 \ cm$

The distance of the hole from the one end $D = 15cm$

From the diagram we see that point A is the center of the brass ring

So the length from the axis of rotation is mathematically evaluated as

$AP = 80 + 10 -5 -15$

$AP = 70 \ cm = (70)/(100) = 0.7 \ m$

Now the period of the pendulum is mathematically represented as

$T = 2 \pi \sqrt{(AP)/(g) }$

$T = 2 \pi \sqrt{(0.7)/(9.8 ) }$

$T = 1.68 \ sec$

Find the current if 20C of charge pass a particular point in a circuit in 10 seconds.

Answers

The current will be "2 A".

Given values are:

Charge, Q = 20 C
Time, t = 10 seconds

As we know,

→ $Current = (Charge)/(Time)$

or,

→ $i = (Q)/(t)$

BY substituting the values, we get

$= (20)/(10)$

$= 2 \ A$

Thus the answer above is right.

Learn more about current here:

brainly.com/question/19668907

Answer:

2 A

Explanation:

From the question,

Q = it..................... Equation 1

Where Q = Quantity of charge, i = cudrrent, t = time.

Make i the subject of the equation

i = Q/t.......................... Equation 2

Given: Q = 20 C, t = 10 seconds.

Substitute these values into equation equation 2

i = 20/10

i = 2 A.

Hence the current is 2A

From her bedroom window a girl drops a water-filled balloon to the ground, 4.77 m below. If the balloon is released from rest, how long is it in the air?

Answers

We need to use the equation x = vt + (1/2)at^2. We know x = 4.77, v = 0, and a = 9.81m/s^2. Plug in the values. 4.77 = (0)t + (1/2)(9.81)t^2 Solve for t. 4.77 = (4.905)t^2 0.972 = t^2 t = (sq.rt)_/0.972 t = 0.985 So it's in the air 0.985 seconds.

A person exerts a horizontal force of F=45N on the end of an 86cm wide door. The magnitude of the torque due to F about the pivot point is determined by |τ|=|rxF|=rFsinθ . Determine the magnitude of the torque, |τ| , on the door about its hinges due to F . |τ|=0Nm |τ|=38.7Nm |τ|=3870Nm

Answers

Answer: The magnitude of torque is 38.7Nm

Explanation: Please see the attachment below

The magnitude of the torque on the door about its h1nges due to the applied force is 38.7 Nm.

How to calculate the magnitude of the torque?

The magnitude of the torque on the door about its h1nges due to the applied force is calculated by applying the following formula as shown below;

τ = rF

where;

r is the perpendicular distance of the applied force
F is the applied force

The given parameters include;

perpendicular distance, r = 86 cm = 0.86 m

the applied force , F = 45 N

The magnitude of the torque on the door about its h1nges due to the applied force is calculated as;

τ = rF

τ = 0.86 m x 45 N

τ = 38.7 Nm

Learn more about torque here: brainly.com/question/30338159

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The presence of dwarf galaxies around the Milky Way supports what picture of our galaxy’s formation?

Answers

Answer:

The presence of dwarf galaxies around the Milky Way supports what picture that our galaxy was formed by a coming together or combination of smaller systems

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A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubber square of the same size. The rubber dielectric is 2.5 mm thick, hasr2.5, andbreakdown field strength of 25 megavolts per meter. Find the voltage rating of the capacitor using a safetyfactor of 10.

A. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 45m?b. How long will it be in the air?

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of 343 m/s.

In an electric circuit, resistance and current are ____A. directly proportional B. inversely proportional C. have no effect on each other