Answer:7.93 m/s
Explanation:
Given
mass of ball 
Mass of hanging ball 
Length of string 
Maximum angle turned 
is the initial velocity of ball 1 and 0 is the initial velocity of ball 2
For Perfectly elastic final velocity of ball 1 and 2 is given by
where 
and 
are the velocity of 1 and 2 before collision
thus 
By energy conservation on second ball we get
Kinetic energy=Potential Energy
thus 
Answer:
The speed of transverse waves in this string is 519.61 m/s.
Explanation:
Given that,
Mass per unit length = 5.00 g/m
Tension = 1350 N
We need to calculate the speed of transverse waves in this string
Using formula of speed of the transverse waves
Where, 
= mass per unit length
T = tension
Put the value into the formula
Hence, The speed of transverse waves in this string is 519.61 m/s.
Answer:
16
Explanation:
solution:
by taking the ratio of the energy E_n,be of the nth orbit of a beryllium atom
Z_be=4 to the energy E_n,h of the nth orbit of a hydrogen (Z_h=1) atom gives
E_n,B/E_n,H=-(2.18*10^-18)*Z^2_BE/-(2.18*10^-18)*Zh^2/n^2
=Z^2_BE+/Z^2_H
=(4)^2/(1)^2
= 16
Answer:
194 kg m²
Explanation:
Use parallel axis theorem:
I = I₀ + mr²
I = 25 kg m² + (1 kg) ((5 m)² + (12 m)²)
I = 25 kg m² + 169 kg m²
I = 194 kg m²
Answer:The Poynting vector SS represents the flow of energy in an EM field. Specifically, if uu is the energy density of the field, the Poynting vector satisfies the continuity equation for it:
∂u∂t+∇⋅S=0
∂u∂t+∇⋅S=0
in vacuum. (This is Poynting's theorem.)
In your particular problem, EE and BB are perpendicular and their cross product is proportional to the product of their amplitudes. Thus
Sz=cμ0B2.
Sz=cμ0B2.
You then have to use your knowledge of BB to work out SS.
Explanation:
