<u>Answer:</u>
The correct answer option is D. The distance between the planet and the Sun changes as the planet orbits the sun.
<u>Explanation:</u>
Kepler’s laws of planetary motion, derived by the German astronomer Johannes Kepler, are the laws of physics that describe the motions of the planets in the solar system.
According to the Kepler's first law of planetary motion: the path on which the planets orbit around the sun is elliptical in shape, with the center of the sun at one focus.
Therefore, the distance between the Sun and the planets vary as the planet orbit around the sun.
Answer:
Resistance =330 Ω
Tolerance = 33 Ω
Explanation:
see attached resistor color code table
The first stripe is orange, which means the leftmost digit is a 3.
The second stripe is orange , which means the next digit is a 3.
The third stripe is brown. Since brown is 1, it means add one zero to the right of the first two digits.
The resistance is:
orange-orange-brown= 330 Ω
The tolerance is:
The fourth color band indicates the resistor's tolerance. Tolerance is the percentage of error in the resistor's resistance.
silver is 10%
A 330 Ω resistor has a silver tolerance band.
<em>Tolerance = value of resistor x value of tolerance band </em>
= 330 Ω x 10% = 33 Ω
330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.
Answer:
Fc = 89.67N
Explanation:
Since the rope is unstretchable, the total length will always be 34m.
From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:
L1+L2=34m
Replacing this value in the previous equation:
Solving for H:

We can now, calculate the angle between L1 and the 2m segment:

If we make a sum of forces in the midpoint of the rope we get:
where T is the tension on the rope and F is the exerted force of 87N.
Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

Answer:
L = 0.635m
Explanation:
This problem involves the concept of stationary waves in pipes. For pipes closed at one end,
The frequency f = nv/4L for n = 1,3,5....n
For pipes open at both ends
f = nv/2L for n = 1,2,3,4...n
Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.
The film solution can be found in the attachment below.