Answer:
The first frequency of audible sound in the speed sound is
f = 662 Hz
Explanation:
vs = 344 m/s
x = 52 cm * 1 / 100m = 0.52m
The wave length is the distance between the peak and peak so
d = 2x
d = 2*0.52 m
d = 1.04 m
So the frequency in the speed velocity is
f = 1 / T
f = vs / x = 344 m/s / 0.52m
f ≅ 662 Hz
Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV
Explanation: To solve this problem we have to use the relationship given by De Broglie as:
λ =p/h where p is the momentum and h the Planck constant
if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m
Finally we have:
eΔV=p^2/2m= h^2/(2*m*λ^2)
replacing we obtained the above values.
Answer:
the answer will be option no b plss mark me brainliest
The height of the ball above the ground is 38.45 m
First we will calculate the velocity of the ball when it touch the ground by using first equation of motion
v=u+gt
v=0+9.81×2.8
v=27.468 m/s
now the height of the ground can be calculated by the formula
v=√2gh
27.468=√2×9.81×h
h=38.45 m
Answer:
F = W + ma a> 0
Explanation:
For this exercise let's use Newton's second law
we assume the upward direction as positive
F - W = m a
F = W + ma
F = m (g + a)
In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive
We can see that since a> 0 the force F must have greater than the weight of the elevator
