Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Answer: Yes the further the sun is away the longer the shadow is. At noon,the shadow is the shortest because its straight up above you. If this helps pls mark brainliest!
Explanation:
the lights on Galaxy watch shifted of uniform amounts to what the rate in of the spectrum,regardless of the distance from Earth
Answer: 1.13 X 10^3 g or 1130 g
Explanation: When you add up everything, you will get 1134.1288 g = 1.11341218 X 10^3. But while adding significant figures, we always look for the least significant figures (here 4.65 has least sig. figures i.e. 3) and take that as a reference to how many significant figures we should have in the answer. (Pretty hard to explain!) Hope you got it!!
Answer:
Explanation:
It is given that,
Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :
R = Rydberg constant, 
Solving above equation we get the value of final n is,
or
So, it will relax in the n = 2. Hence, this is the required solution.
