Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
Answer:
Explanation:
The path length difference = extra distance traveled
The destructive interference condition is:
![\Delta d = (m+1/2)\lambda](https://tex.z-dn.net/?f=%5CDelta%20d%20%3D%20%28m%2B1%2F2%29%5Clambda)
where m =0,1, 2,3........
So, ←
![\Delta d = (m+1/2)\lamb da9/tex]so [tex]\Delta d = \frac{\lambda}{2}](https://tex.z-dn.net/?f=%5CDelta%20d%20%3D%20%28m%2B1%2F2%29%5Clamb%20da9%2Ftex%5D%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3Eso%20%3C%2Fstrong%3E%5Btex%5D%5CDelta%20d%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%7D)
⇒ λ = 2Δd = 2×10 = 20
Energy of a random atomic and molecular is called molecules energy
Answer:
<h3>What is the angular speed of the earth around the sun? </h3>
It takes the Earth approximately 23 hours, 56 minutes and 4.09 seconds to make one complete revolution (360 degrees). This length of time is known as a sidereal day. The Earth rotates at a moderate angular velocity of
![7.2921159 × 10 {}^{ - 5} \: \: \frac{radians}{second}](https://tex.z-dn.net/?f=7.2921159%20%C3%97%2010%20%7B%7D%5E%7B%20-%205%7D%20%20%5C%3A%20%20%5C%3A%20%20%5Cfrac%7Bradians%7D%7Bsecond%7D%20)
<h3>
What is the tangential speed of the earth? </h3>
The earth rotates once every 23 hours, 56 minutes and 4.09053 seconds, called the sidereal period, and its circumference is roughly 40,075 kilometers. Thus, the surface of the earth at the equator moves at a speed of 460 meters per second--or roughly 1,000 miles per hour.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The theoretical angular magnification lies within the angular magnification range
Explanation:
From the question we are told that
The focal length of B is ![f_{objective } = 43.0 \ cm](https://tex.z-dn.net/?f=f_%7Bobjective%20%7D%20%3D%20%2043.0%20%5C%20cm)
The focal length of A is ![f_{eye} = 10.4 \ cm](https://tex.z-dn.net/?f=f_%7Beye%7D%20%3D%20%2010.4%20%5C%20%20cm)
The theoretical angular magnification is mathematically represented as
![m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Bf_%7Bobjective%20%7D%7D%7Bf_%7Beye%7D%7D%20%20%3D%20%20%5Cfrac%7B43.0%7D%7B10.4%7D)
![m = \frac{f_{objective }}{f_{eye}} = 4.175](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Bf_%7Bobjective%20%7D%7D%7Bf_%7Beye%7D%7D%20%20%3D%20%204.175)
Form the question the measured angular magnification ranges from 4 -5
So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range