To answer that question, we don't care what the highest and lowest
levels of the wave are, or how far apart they are. We only need to be
able to identify the highest point on the wave, and keep track of how
often those pass by us.
You said it takes 4 seconds for a complete wave to pass by.
Through the sheer power of intellect, I'm able to take that information
and calculate that 1/4 of the wave passes by in 1 second.
There's your frequency . . . 1/4 per second, or 0.25 Hz.
Answer:
Explanation:
From the question we are told that
Weight 
Altitude 
Speed 
Generally the equation for Potential energy ids mathematically given as
<h2>
Answer:</h2>
1000th multiple of the standard reference level for intensities.
<h2>
Explanation:</h2>
The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;
β = 10 log (I / I₀) --------------------(i)
Where;
I₀ = reference intensity
Given from the question;
β = sound level = 30dB
Substitute this value into equation (i) as follows;
30 = 10 log (I / I₀)
Divide both sides by 3;
3 = log (I / I₀)
Take antilog of both sides;
10^(3) = (I / I₀)
1000 = I / I₀
Solve for I;
I = 1000I₀
Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)
Answer:
W = - 5.01 10¹⁰ J
Explanation:
Work is defined by the expression
W = ∫ F.dr
Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product
W = ∫ F dr
W = G m₁m₂ ∫ 1 /r² dr
W = G m₁ m₂2(-1 / r)
We evaluate between the lower limits r = Re and upper r = ∞
W = G m₁m₂ (-1 / Re + 1 / ∞)
W = - G m₁ m₂ / Re
Let's calculate
W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶
W = - 5.01 10¹⁰ J
