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rosijanka [135]
4 years ago
13

What happens to incoming light rays that are parallel to the principal axis of a convex lens? A. They converge at the focal poin

t. B. They diverge away from the principal axis. C. They converge beyond the focal point. DO. They diverge from the focal point.
Physics
2 answers:
kaheart [24]4 years ago
5 0

Answer:

They diverge from the focal point.

Explanation:

A convex lens is also called as converging lens. If the rays of light are parallel to the principal axis of a convex lens, it passes through the focal point of the lens.

For a converging lens, the ray of light diverge from the focal point of the lens. The image formed by a convex lens is real and inverted. When the object is placed between the focus and the optical center, the image is virtual and erect.

So, the correct option is (d) "They diverge from the focal point".

snow_tiger [21]4 years ago
3 0

Answer:

D. They diverge from the focal point.

Explanation:

Incoming light rays diverge from the focal point is they are parallel to the principal axis of a convex lens.

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It will rise, and then fall back to the Earth’s surface.
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Which statement best compares potential and kinetic energy?
xenn [34]

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D

Explanation:

Because kinetic energy do have more than potential energy: kinetic energy is when a object is moving. Potential energy is when something is at rest and has no movement what so ever

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Match the following words with their definitions.
Brrunno [24]

Answer:

1. Doppler Effect = The apparent change in frequency perceived when the source of sound is approaching or moving away from an observer

2. Beats = The vibration of an object at its natural frequency, caused by a force or wave of the same frequency

3. Resonance = a series of alternating reinforcements and cancellations produced by the interference of waves

4. Shockwave = The cone-shaped wave made by a particle moving faster than the speed of a wave in that medium

Explanation:

I hope this helps. :)

7 0
4 years ago
Read 2 more answers
A 3.70 kg block moving 2.09 m/s right hits a 2.42 kg block moving 3.92 m/s left. Find the total momentum of the system.
Novosadov [1.4K]

Answer:

-1.7534 Kg.m/s

Explanation:

Momentum, p is a product of mass and velocity, expressed as p= mv where m is the mass and v is the velocity which is dependent on the direction. Taking left as negative and right as positive then the initial momentum will be

p=-2.42*3.92+(3.7*2.09)=-1.7534 Kg.m/s

5 0
3 years ago
Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F
muminat

Answer:

v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)

Explanation:

Given that the force of the particle is,

F_{x}=F_{0}e^{-kt}

Now it can be further written as

m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C

Now the initial conditions are v=1 at t=0.

So,

1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}

Now the velocity will become.

v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

And,

\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\

And, another initial condition is x=0 at t=0

0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}

Now,

x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)

5 0
3 years ago
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