Answer:
The Magnifying power of a telescope is ![M = 109.26](https://tex.z-dn.net/?f=M%20%3D%20109.26)
Explanation:
Radius of curvature R = 5.9 m = 590 cm
focal length of objective
= ![\frac{R}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B2%7D)
⇒
= ![\frac{590}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B590%7D%7B2%7D)
⇒
= 295 cm
Focal length of eyepiece
= 2.7 cm
Magnifying power of a telescope is given by,
![M = \frac{f_{objective} }{f_{eyepiece} }](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7Bf_%7Bobjective%7D%20%7D%7Bf_%7Beyepiece%7D%20%7D)
![M = \frac{295}{2.7}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7B295%7D%7B2.7%7D)
![M = 109.26](https://tex.z-dn.net/?f=M%20%3D%20109.26)
therefore the Magnifying power of a telescope is ![M = 109.26](https://tex.z-dn.net/?f=M%20%3D%20109.26)
A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be 372.77 m/s
<h3>What is a sound wave?</h3>
It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.
For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula
V= √F/μ
where v is the wave velocity of the wave travel on the string
F is the tension in the string of piano
μ is the mass per unit length of the string
As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.
The μ is the mass per unit length of the string would be
μ = 4.80/(1.150×1000)
μ = 0.0041739 kg/m
By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity
V= √F/μ
V=√(580/0.0041739)
V = 372.77 m/s
Thus, the speed of transverse waves on the wire comes out to be 372.77 m/s
Learn more about sound waves from here
brainly.com/question/11797560
#SPJ1
Answer:
88.3
Explanation:
Emf in a rotating coil is given by rate of change of flux:
E= dФ/dt=(NABcos∅)/ dt
N: number of turns in the coil= 80
A: area of the coil= 0.25×0.40= 0.1
B: magnetic field strength= 1.1
Ф: angle of rotation= 90- 37= 53
dt= 0.06s
E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V
I believe the answer is californium because it is used in most metals and is very strong and expensive like gold and silver