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3 years ago

The lens in the eyepiece of a telescope magnifies light by bending it. What is this bending called?

Physics

2 answers:

PtichkaEL [24]3 years ago

8 0

Answer:

It is called refraction.

Explanation:

When a ray of light enters through one medium to another, it deviates from its path. This deviation is called refraction.

When ray travel from rarer medium to denser medium then after refraction, it bends towards the normal.

When ray travels from denser medium to rarer medium, it bends away from the normal.

tresset_1 [31]3 years ago

7 0

I believe that it is called refracting...

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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.

Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

The given parameters;

speed of electron, v = 2.2 x 10⁶ m/s
radius of the circle, r = 4.63 x 10⁻¹¹ m

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

$F = \frac{M_e v^2}{r} \\\\$

where;

$M_e$ is mass of electron = 9.11 x 10⁻³¹ kg

$F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N$

Thus, the central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

Learn more about centripetal force here:brainly.com/question/20905151

8 0

3 years ago

How does changes in distance affect the gravitational pull between two objects? Describe and give one example.

maxonik [38]

The formula is

F_grav = G * m1 * m2 / r^2

G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.

As the distance increases, the Gravitational Force will decrease. It will decrease by quite a bit.

As the distance decreases, the gravitational force will Increase.

The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.

8 0

4 years ago

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

Strike441 [17]

Answer:

θ = 13.7º

Explanation:

According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

$\Delta K + \Delta U = W_{nc} (1)$

ΔK, is the change in kinetic energy, as follows:

$\Delta K = \frac{1}{2}* m* (v_{f} ^{2} - v_{0} ^{2}) (2)$

ΔU, is the change in the gravitational potential energy.
If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º

8 0

3 years ago

A single conservative force acts on a 5.30-kg particle within a system due to its interaction with the rest of the system. The e

cupoosta [38]

Answer:

Given that

m = 5.3 kg

Fx = 2x + 4

We know that work done by force F given as

w= ∫ F. dx

Given that x=1.08 m to x=6.5 m

Fx = 2x + 4

w= ∫ F. dx

$w=\int_{1.08}^{6.5}(2x+4) .dx$

$w=\left [x^2+4x \right ]_{1.08}^{6.5}$

$w=(6.5^2-1.08^2)+4(6.5-1.08)\ J$

w=62.7 J

We know that potential energy given as

$F=-\dfrac{dU}{dx}$

∫ dU = -∫F.dx ( w= ∫ F. dx)

ΔU= -62.7 J

We know that form work power energy theorem

Net work = Change in kinetic energy

W= KE₂ - KE₁

62.7 =KE₂ - (1/2)x 5.3 x 3²

KE₂ = 86.55 J

This is the kinetic energy at 6.5m

8 0

3 years ago

Consider the two vectors A = 3 î − ĵ and B = − î − 2 ĵ.

LUCKY_DIMON [66]

Answer:

(a) A+B = 2i-3j

(B) A-B = 4i + j

Explanation:

We have given two vectors A = 3i-j and B = -1-2j

We have to find the two vectors that is A+B and A-B

(A) In first art we have calculate A+B for this we have to add simply vector A and v ector B

So A+B = 3i-j-i-2j = 2i-3j

(B) In this part we have to find A-B for this we have to simply subtract B from A so A-B = 3i-j-(-i-2j) =3i-j+i+2j =4i+j

6 0

3 years ago