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lesya [120]
3 years ago
10

What happens to parallel light rays that strike a concave lens?

Physics
1 answer:
torisob [31]3 years ago
5 0

Answer:

They diverge on refraction

Explanation:

When parallel light rays strike a concave lens, they will diverge that is they spread out .

Concave lens is also known as diverging lens, which means that when parallel rays of light strike on it, the lens spreads out the light rays ( that is it diverges the rays of light) that are refracted through it.

At the middle  of concave lens is thinner.  

When light is passes through the lens they diverge it or spread out.

The concave lens causes light rays to bend away or diverge from its axis since the concave lens is a diverging lens.  

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Distance is how far something moves in a specific amount of time. 
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1) A change in momentum is the result of an
dimaraw [331]

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Explanation: Since the equation for momentum is p (momentum) = m (mass)* v (velocity), if the mass or velocity (or both) of the object is changed, the momentum will change.

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1.What type of sediment forms from minerals that crystallize from seawater?
nikdorinn [45]
The correct answers are as follows:
<span>1) hydrogenous sediment

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3 years ago
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A medicine ball has a mass of 6 kg and is thrown with a speed of 4 m/s. What is its kinetic energy?
zhenek [66]

Answer:

\boxed{\sf Kinetic \ energy \ (KE) = 48 \ J}

Given:

Mass (m) = 6 kg

Speed (v) = 4 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 6 \times  {4}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 3 \times 16

\sf \implies KE =3 \times 16

\sf \implies KE = 48 \: J

6 0
4 years ago
A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang
Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]

E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]

Since \lambda = \frac{Q}{L}

where;

L = length

Q = charge

λ =  density of the charge;

then substituting \frac{Q}{L} for λ, we have :

E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]

E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}

substituting our given parameter; we have:

E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0
3 years ago
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