Ask question

Ask question

All categories

katen-ka-za [31]

3 years ago

5

A refracting telescope has a 1.48 m diameter objective lens with focal length 15.4 m and an eyepiece with focal length 3.28 cm.

What is the angular magnification of the telescope

1 answer:

vivado [14]3 years ago

7 0

Answer:

m = 469.51

Explanation:

Given that,

The diameter of a refracting telescope is 1.48 m

Focal length of the objective lens is 15.4 m

Focal length of an eyepiece is 3.28 cm

We need to find the angular magnification of the telescope. The ratio of focal length of objective lens to the focal length of the eye piece is called angular magnification of the telescope. So,

$m=\dfrac{f_o}{f_e}\\\\m=\dfrac{15.4}{3.28\times 10^{-2}}\\\\m=469.51$

So, the angular magnification of the telescope is 469.51.

You might be interested in

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (60pts)

mr Goodwill [35]

The density of an object determines whether it will float or sink in another substance. An object will float if it is less dense than the liquid it is placed in. An object will sink if it is more dense than the liquid it is placed in.
So since the boat has a lower density than the water, it will float.
So the answer is choice B

3 0

2 years ago

an elevator mass of 7700 kg falls from a height of 32 m after a sudden failure in the hoisting cable. The mass is stopped by a s

valkas [14]

Answer:k=28.29 kN/m

Explanation:

Given

mass $m =7700 kg$

height from which Elevator falls $h=32 m$

Let x be the compression in the spring

thus From conservation of Energy Potential energy will convert in to Elastic Potential Energy of spring

$\frac{kx^2}{2}=mg(h+x)$ ----------1

also maximum acceleration is 5g

thus

$mg-kx=ma$

here $a=-5g$

$kx=mg-m(-5g)=6mg$

$x=\frac{6mg}{k}$

Substitute x in equation 1

$0.5\times k\times (\frac{6mg}{k})^2=mg(h+\frac{6mg}{k})$

$18\frac{(mg)^2}{k}=mgh+6\frac{(mg)^2}{k}$

$k=12\cdot \frac{mg}{h}$

$k=12\times \frac{7700\times 9.8}{32}$

$k=28.29 kN/m$

4 0

2 years ago

A fast-food restaurant uses a conveyor belt to send the burgers through a grilling machine. if the grilling machine is 1.2 m lon

Vinil7 [7]

length of the grilling machine is 1.2 m

time taken to cook the burger is 2.7 min = 162 s

so the speed of the machine should be like this that if must have to cook till it cross the machine

$v = \frac{d}{t}$

$v = \frac{1.2}{162}$

$v = 7.41* 10^{-3} m/s$

now in one minute the total length of the machine that is covered is given by

$L = v*t$

$L = 7.41*10^{-3}* 60 = 44.4 cm$

now distance between the burgers is 15 cm

so total production rate will be

$N = \frac{44.4}{15} = 3 burger/min$

so it will produce 3 burger per minute

8 0

2 years ago

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotr

mart [117]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v² ........1

6.5 × 1.6 × $10^{-19}$ = 1/2 × 1.672 × $10^{-27}$ ×v²

v = 3.5 × $10^{7}$ m/s

so

radius will be

radius = $\frac{m*v}{B*q}$ ........2

radius = $\frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}$

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

8 0

2 years ago

Read 2 more answers

A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its or

Brut [27]

A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its original wavelength. The sound wave traveled through a helium balloon (helium is less dense than air could explain this change in wavelength

The pattern of disruption brought on by energy moving away from the sound source is known as a sound wave. Longitudinal waves are what makeup sound. This indicates that the direction of energy wave propagation and particle vibrational propagation are parallel. The atoms oscillate when they are put into vibration.

A high-pressure and a low-pressure zone are created in the medium as a result of this constant back and forth action. Compressions and rarefactions, respectively, are terms used to describe these high- and low-pressure zones. The sound waves go from one medium to another as a result of these regions being transmitted to the surrounding media.

To learn more about sound waves please visit -
brainly.com/question/11797560
#SPJ1

4 0

1 year ago