Answer:
W = 0.842 J
Explanation:
To solve this exercise we can use the relationship between work and kinetic energy
W = ΔK
In this case the kinetic energy at point A is zero since the system is stopped
W = K_f (1)
now let's use conservation of energy
starting point. Highest point A
Em₀ = U = m g h
Final point. Lowest point B
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
mg h = K
to find the height let's use trigonometry
at point A
cos 35 = x / L
x = L cos 35
so at the height is
h = L - L cos 35
h = L (1-cos 35)
we substitute
K = m g L (1 -cos 35)
we substitute in equation 1
W = m g L (1 -cos 35)
let's calculate
W = 0.500 9.8 0.950 (1 - cos 35)
W = 0.842 J
Yes, in general, metalloids are slightly reactive.
Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.
But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.
From the first law of thermodynamics, we use the equation expressed as:
ΔH = Q + W
where Q is the heat absorbed of the system and W is the work done.
We calculate as follows:
ΔH = Q + W
ΔH = 829 J + 690 J = 1519 J
Hope this answers the question. Have a nice day.
Explanation:
The given data is as follows.
Length (l) = 2.4 m
Frequency (f) = 567 Hz
Formula to calculate the speed of a transverse wave is as follows.
f = 
Putting the gicven values into the above formula as follows.
f =
567 Hz = 
v = 544.32 m/s
Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.
