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3 years ago

What happens to light when it passes through a lens?

2 answers:

7 0

B. Light refracts as it passes through a lens.

The bending of a ray of light also occurs when light passes into and out of a glass lens. ... Because a convex lens can cause rays of light to converge, it can produce an image on a screen.

Art [367]3 years ago

5 0

Answer:B the light refracts.

Explanation: a good example is an straw in a cup. When the light goes through the glass the straw looks broken an bigger

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A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface a

kykrilka [37]

Answer:

v₀ = 280.6 m / s

Explanation:

we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,

We write the mechanical energy when the shock has passed the bodies

Em₀ = K = ½ (m + M) v²

We write the mechanical energy when the spring is in maximum compression

$Em_{f} = K_{e} \\= \frac{1}{2} kx^2\\ Em_0 = Em_{f}$

½ (m + M) v² = ½ k x²

Let's calculate the system speed

v = √ [k x² / (m + M)]

v = √[152 ×0.78² / (0.012 +0.109) ]

v = 27.65 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

p₀ = m v₀

After the crash

$p_{f} = (m + M) v$

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

$p_0 = p_{f}$

m v₀ = (m + M) v

v₀ = v (m + M) / m

let's calculate

v₀ = 27.83 (0.012 +0.109) /0.012

v₀ = 280.6 m / s

4 0

3 years ago

What happens when you place two magnets close together.

wlad13 [49]

They attract and stick together

3 0

2 years ago

nsider a common mirage formed by superheated air immediately above a roadway. A truck driver whose eyes are 2.00 m above the roa

Anna11 [10]

Answer:

$n_{air} =1.0000355$

Explanation:

from snells law we have following relation

$n' Sin i = n_{air} Sin r$

$
n_{air}$ is refractive index of air

n' - refractive index of air below

i is angle of incidence = 90 - 1.30 = 88.70^o

r is refraction angle

plugging all value in above formula

$1.000293 \times sin 88.7 = n_{air} \times sin 90$

$n_{air} =1.0000355$

6 0

3 years ago

QUESTION: A study group of students is reviewing claims in an article that gamma rays are among the most dangerous type of elect

Lubov Fominskaja [6]

Bias is an inclination towards one way of thinking due to inherent factors in our environment, such as how one was brought up.

Explanation:

In science, biases, conscious or unconscious, tend to make experiments’ results to incline in particular way rather than being impartial. Peer- review of scientific journals by the scientific community is meant to be a quality assurance measure to detect such biases. The students should check reviews of the scientific community on the article and see if any biases have been pointed out. Probably the institution that the scientist works for is in competition with another institution that deals with gamma rays.

Learn More:

brainly.com/question/1903490

brainly.com/question/8480257

#LearnWithBrainly

4 0

3 years ago

Read 2 more answers

The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =

dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar, θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

∑ Mₐ = Iα

Where, Mₐ - the moment about A

α - angular acceleration

I - moment of inertia of the rod AB

$-mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha$

$\alpha=\frac{-3gcos\theta}{2l}$

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

$\vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

The acceleration at the center of the bar

$\vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}$

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]$