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2 years ago

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If the focal length of a reflection telescope is 200 cm and the focal length of the eyepiece lens is 0.25 cm, what is the magnif

ying power of the telescope?

2 answers:

Serga [27]2 years ago

7 0

Answer:

-48cm

Explanation:

the following data are given

focal length of telescope=200cm,

focal length of the eyepiece=0.25cm

From the genera formula used to find the magnifying power which is expressed as

$M=-\frac{fx_{o}}{f_{e}}[1+\frac{f_{e}}{d}]$

where

$f_{e} = focal length of thr eye piece\\ f_{o} =focal length of the telescope\\$

and d=least distance of distinct vision=25cm

if we substitute values into the formula, we arrive at

$M=-\frac{fx_{o}}{f_{e}}[1+\frac{f_{e}}{d}]\\M=-\frac{200cm}{0.25cm}[1+\frac{0.25cm}{25cm}]\\M=-800cm[1+0.01]\\M=-800cm(1.01)\\M=-808cm \\M=-808cm$

hence from the answer, we can conclude that the magnifying power of the telescope is -808cm

masha68 [24]2 years ago

5 0

Answer:

800

Explanation:

Focal length of telescope, F = 200cm

Focal length of eyepiece, f = 0.25

The magnifying power of a telescope is given as the ratio of the focal length of the objective of the telescope to the focal length of the lens. Mathematically:

M = F/f

Therefore, when F = 200cm and f = 0.25cm:

M = 200/0.25

M = 800

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A small rivet connecting two pieces of sheet metal is being clinched by hammering. Determine the impulse exerted on the rivet an

kykrilka [37]

Answer:

a) the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b) the impulse exerted by the rivet when the anvil has a support weight of 9 lb = 0.799 lb.s

the energy absorbed by the rivet under each blow when the anvil has a support weight of 9 lb is = 7.99 ft.lb

Explanation:

The first picture shows a schematic view of a free body momentum diagram of the hammer head and the anvil.

Using the principle of conservation of momentum to determine the final velocity of anvil and hammer after the impact; we have:

$m_Hv_H + m_Av_A = m_Hv_2+m_Av_2$

From the question given, we can deduce that the anvil is at rest;

∴ $v_A = 0$ ; then, we have:

$m_Hv_H + 0 = (m_H+m_A) v_2$

Making $v_2$ the subject of the formula; we have:

$v_2 = \frac{m_Hv_H}{m_H + m_A}$ ------- Equation (1)

Also, from the second diagram; there is a representation of a free body momentum of the hammer head;

From the diagram;

F = impulsive force exerted on the rivet

Δt = the change in time of application of the impulsive force

Using the principle of impulse of momentum to the hammer in the quest to determine the impulse exerted (i.e FΔt ) on the rivet; we have:

$m_Hv_H - F \delta t = m_Hv_2$

$- F \delta t = - m_Hv_H + m_Hv_2$

$F \delta t = m_Hv_H - m_Hv_2$

$F \delta t = m_H(v_H - v_2)$ ------- Equation (2)

Using the function of the kinetic energy of the hammer before impact $T_1$ ; we have:

$T_1 = \frac{1}{2} m_Hv_H^2$ -------- Equation (3)

We determine the mass of the hammer $m_H$ by using the formula from:

$W_H = m_Hg$

where;

$W_H$ = weight of the hammer

$m_H$ = mass of the hammer

g = acceleration due to gravity

Making $m_H$ the subject of the formula; we have:

$m_H = \frac{W_H}{g}$

$m_H = \frac{1.5 \ lb}{32.2 \ ft/s^2}$

$m_H = 0.04658 \ lb.s^2/ft$

Now;

$T_1 = \frac{1}{2} m_Hv_H^2$

$T_1 = \frac{1}{2}*(0.04658 \ lb.s^2 /ft) *(20 \ ft/s)^2$

$T_1 = \frac{18.632 }{2}$

$T_1 = 9.316 \ ft.lb$

After the impact $T_2$ ; the final kinetic energy of the hammer and anvil can be written as:

$T_2 = \frac{1}{2}(m_H +m_A)v^2_2$

Recall from equation (1) ; where $v_2 = (\frac{m_Hv_H}{m_H+m_A})$ ; if we slot that into the above equation; we have:

$T_2 = \frac{1}{2}(m_H +m_A)( \frac{m_Hv_H}{m_H+m_A})^2$

$T_2 = \frac{1}{2} \frac{m^2_H +v^2}{m_H+m_A}$

$T_2 = \frac{1}{2} ({m^2_H +v^2})(\frac{m_H}{m_H+m_A})$

Also; from equation (3)

$T_1 = \frac{1}{2} m_Hv_H^2$ ; Therefore;

$T_2 = T_1 (\frac{m_H}{m_H+m_A})$ ----- Equation (4)

a)

Now; To calculate the impulse exerted by the rivet FΔt and the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support; we have the following process

First , we need to find the mass of the anvil when we have an infinite mass support;

mass of the anvil $m_A = \frac{W_A}{g}$

where we replace; $W_A \ with \ \infty$ and g = 32.2 ft/s²

$m_A = \frac{\infty}{32.2 \ ft/s}$

However ; from equation (1)

$v_2 = \frac{m_H v_H}{m_H + m_A}$

$v_2 = \frac{0.04658*20}{0.04658+ \ \infty}$

$v_2 = 0$

From equation (2)

$F \delta t = m_H(v_H + v_2)$

$F \delta t = (0.04658 lb .s^2 /ft )(20ft/s - 0)$

$F \delta t = \ 0.932 \ lb.s$

Therefore the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

For the energy absorbed by the rivet ; we have:

$T_2 = T_1 (\frac{m_H}{m_H+m_A} )$

where;

$T_1= 9.316 \ ft.lb$

$m_H = 0.04658 \ lb.s^2/ft$

$m_A = \infty$

Then;

$T_2 = (9.316 \ ft.lb) (\frac{0.04658\ lb.s^2/ft)}{0.04658 \ lb.s^2/ft+ \infty} )$

$T_2 = (9.316 \ ft.lb)* 0$

$T_2 = 0$

Then the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support

ΔT = $T_1 - T_2$

ΔT = 9.316 ft.lb - 0

ΔT ≅ 9.32 ft.lb

Therefore; we conclude that the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b)

Due to the broadness of this question, the text is more than 5000 characters, so i was unable to submit it after typing it . In the bid to curb that ; i create a document for the answer for the part b of this question.

The attached file can be found below.

5 0

2 years ago

The wavelength of green light is 550 nm.

SIZIF [17.4K]

Answer:

(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s

velocity of electron is 1323.88 m/s

momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

velocity of electron is 1654.85 m/s

momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1323.88)

p = 1.205 x 10⁻²⁷ kgm/s

(b)

wavelength of red light, λ = 440 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{440*10^{-9}}\\\\p = 1.506 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(440*10^{-9})}\\\\v = 1654.85 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1654.85)

p = 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron.

7 0

2 years ago

In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired

Sauron [17]

They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)

4 0

2 years ago

A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw

navik [9.2K]

Answer: 75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7 x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 = 200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

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2 years ago

Read 2 more answers

How can controlling the way light bends and reflects be used to help people?

Vinil7 [7]

too much sun is dangerous for humans and can cause cancer so it's important that light is reflected for example a pool reflects water back to space that is why water sometimes is cold because it reflects light

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2 years ago