Answer:
Explanation:
Given:
- relativistic length of stick A,
- relativistic velocity of stick A with respect to observer,
<em>Since the object is moving with a velocity comparable to the velocity of light with respect to the observer therefore the length will appear shorter according to the theory of relativity.</em>
<u> Mathematical expression of the theory of relativity for length contraction:</u>
where:
L = relativistic length
original length at rest
Lorentz factor 
Answer:
<u>Frequency</u>- number of wave cycles that occur in a given amount of time.
<u>Pitch</u>- number of wavelengths in a given amount of time.
<u>Amplitude</u>- fluctuation or displacement of a wave from its mean value. That means how high or low they are away from the center line.
<u>Volume</u>- The perception of loudness from the intensity of a sound wave. The higher the intensity of a sound, the louder it is perceived in our ears, and the higher volume it has.
<u>Wavelength</u>- the distance between the tops of the "waves".
<span>11.823 cm
There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall.
The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm.
Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So
82 + 1/21 * 2 = 82 + 2/21 = 82.0952381
Now we have the following dimensions with a circle replacing the ball in the original problem.
Distance from wall to effective circle = 82.0952381 cm
Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm
Effective diameter of circle = 3.995462279 cm
And because the geometry makes similar triangles, the following ratio applies.
3.995462279/41.9047619 = X/124
Now solve for X
3.995462279/41.9047619 = X/124
124*3.995462279/41.9047619 = X
495.4373226/41.9047619 = X
11.82293611 = X
The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
I like to just keep writing them over and over on a page also if you remember them just before the exam than as soon as you start write them on the front of your test so you don’t forget them.
Hope this helps.
Answer:
19320 K
Explanation:
The temperature of a star is related to its peak wavelength by Wien's displacement law:
where
T is the absolute temperature at the star's surface
is Wien's displacement constant
is the peak wavelength
Here we have
Substituting into the equation, we find
