PHYSICS COLLEGE

What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.5 mm and an eyepiece whose focal length is 2.9 cmcm ? Follow the sign conventions.

Answers

Answer 1

Answer:

The magnifying power of an astronomical telescope will be:

"0.095".

Telescope: Focal length and Power

According to the question,

Radius of curvature, R = 5.5 mm

Focal length of eyepiece, $F_e$ = 2.9 cm

We know that,

→ Focal length of mirror,

F₀ = $(Radius \ of \ curvature)/(2)$

By substituting the values,

= $(5.5)/(2)$

= 2.75 mm or,

= 0.278 cm

hence,

The telescope's magnification be:

= $(F_0)/(F_e)$

= $(0.275)/(2.9)$

= 0.095

Thus the above approach is correct.

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Answer 2

Answer:

0.095

Explanation:

An astronomical telescope is a telescope used for viewing far distance object. It uses two lenses called the objective lens and the eyepiece lens.

Each lens has its own focal length

Let the focal length of the objective lens be Fo

Focal length of the eyepiece be Fe

Magnification of an astronomical telescope = Fo/Fe

Since the telescope uses a reflecting mirror having radius of curvature of 5.5mm instead of an objective lens, then we will replace Fo as the focal length of the mirror.

Focal length of a mirror Fo = Radius of curvature/2

Fo = R/2

Fo = 5.5/2

Fo = 2.75mm

Converting 2.75mm to cm gives 2.75/10 = 0.275cm

Fo = 0.275cm; Fe = 2.9cm

Magnification of the telescope = 0.275/2.9

Magnification of the astronomical telescope = 0.095

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An infant throws 7 g of applesauce at a velocity of 0.5 m/s. All of the applesauce collides with a nearby wall and sticks to it. What is the decrease in kinetic energy of the applesauce?

Answers

Answer:

Δ KE = - 8.75 x 10⁻⁴ J

Explanation:

given,

mass of applesauce = 7 g = 0.007 Kg

initial velocity, u = 0.5 m/s

final velocity, v = 0 m/s

Decrease in kinetic energy = ?

initial kinetic energy

$KE_1=(1)/(2)mu^2$

$KE_1=(1)/(2)* 0.007 * 0.5^2$

KE₁ = 8.75 x 10⁻⁴ J

final kinetic energy

$KE_2=(1)/(2)mv^2$

$KE_2=(1)/(2)* 0.007 * 0^2$

KE₂ =0 J

Decrease in kinetic energy

Δ KE = KE₂ - KE₁

Δ KE = 0 - 8.75 x 10⁻⁴

Δ KE = - 8.75 x 10⁻⁴ J

decrease in kinetic energy of the applesauce is equal to 8.75 x 10⁻⁴ J

Final answer:

The decrease in kinetic energy of the applesauce, when it hits the wall and stops, is the initial kinetic energy of it. Using the formula of kinetic energy, the decrease is calculated to be 0.000875 Joules.

Explanation:

This question relates to the concept of kinetic energy in physics. Kinetic energy is calculated by the formula 0.5 * mass (kg) * velocity (m/s)^2. So the initial kinetic energy of the applesauce right after being thrown was 0.5 * 0.007 kg * (0.5 m/s)^2 = 0.000875 Joules.

When the applesauce hits the wall and stops, its velocity drops to 0. Thus, its kinetic energy also goes to 0 (because kinetic energy is proportional to the square of velocity).

Therefore, the decrease in kinetic energy is the same as the initial kinetic energy of the applesauce, which is 0.000875 Joules.

Learn more about Kinetic Energy here:

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PLEASE HELP DUE BEFORE 11:30 TODAY!!!!Which of the following quantities is NOT a vector quantity?
A. 926 m to the north
B. 5.2 m/s to the west
C. 46 m down
D. 12.3 m/s faster

Answers

Answer:

D is not the a vector quantities

The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00o , what are the underwater angles of refraction for the blue and red components of the light

Answers

Answer:

The underwater angles of refraction for the blue and red components of the light is 47.8° and 48.2°

Explanation:

Using the Snell's law

n1 * sin Ф1 = n2 sin Ф2

1 * sin 83 = n2 sin Ф2

Ф2 = $sin^(-1) ((1)/(n2) * sin 83)$

Red light

n2 = 1.331

Ф2 = $sin^(-1) ((1)/(1.331) * sin 83) = 48.2$ °

Blue light

n2 = 1.340

Ф2 = $sin^(-1) ((1)/(1.340) * sin 83) = 47.8$ °

Match each planet to an accurate characteristic

Answers

Answer:

venus - 2

earth - 3

mars - 4

mercury - 1

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Answers

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

$E_i=(mv^2)/(2)+mgh$

Here , initial velocity is zero .

Therefore , $E_i=mgh=40\ J$

Now , final energy of the system :

$E_f=(mv^2)/(2)+mg(0)+15\n\nE_f=(0.05* v^2)/(2)+15$

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

$E_i=E_f\n40=(0.05v^2)/(2)+15\n\n25=(0.05v^2)/(2)\n\nv=31.62\ m/s$

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

Final answer:

Using the principle of conservation of energy, the speed of the ball just before hitting the Earth's surface is found to be 79.2 m/s after accounting for the 15 J increase in thermal energy.

Explanation:

This question is concerned with the concept of conservation of energy, specifically the principles of potential and kinetic energy. When the ball is 80 meters above the Earth's surface, the total gravitational potential energy is m*g*h = 50g*9.8m/s²*80m = 39200 J (where m is mass, g is gravity, and h is height), and the kinetic energy is 0.

As the ball falls, its potential energy gets converted into kinetic energy, but we also know that the total thermal energy of the ball and the air in the system increases by 15 J. That means that not all the potential energy is converted into kinetic energy, 15 J is lost to thermal energy. So, the kinetic energy of the ball when it hits the Earth is 39200 J - 15 J = 39185 J.

Finally, we know that kinetic energy equals (1/2)*m*v², where v is the speed of the ball. Rearranging this formula to solve for v we get, v = sqrt((2*kinetic energy)/m) = sqrt((2*39185 J)/50g) = 79.2 m/s. So, just before the ball hits the surface, its speed is 79.2 m/s.

Learn more about Potential and Kinetic Energy here:

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A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 76.2-kg water-skier has an initial speed of 5.0 m/s. Later, the speed increases to 10.4 m/s. Determine the work done by the net external force acting on the skier.