Answer:
, the minus meaning west.
Explanation:
We know that linear momentum must be conserved, so it will be the same before (
) and after (
) the explosion. We will take the east direction as positive.
Before the explosion we have 
.
After the explosion we have pieces 1 and 2, so 
.
These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.
Since we know momentum must be conserved we have:
Which means (since we want 
and 
):
So for our values we have:
Answer:
i = 0.00077A
Explanation:
Given:
loop radius, r = 3.0 cm = 0.03 m
Area, A = π x r² = π x 0.03² = 0.0028 m²
Magnetic Field, B = 0.75 T
Loop resistance, R = 18 Ω
time, t = 0.15 seconds
Now,
the induced emf is given as:
EMF = - BA/t .......1
Likewise,
EMF = iR.......2
Equate 1 and 2
iR = - BA/t
i = - BA/tR
i = 0.75×0.0028/0.15×18
i = 0.0021/2.7
i = 0.00077A
Answer:
* in a tube with both ends open
λ = 2L / n n = 1, 2, 3,…
* tube with one end open and the other closed.
λ = 4L / n n = (2n ’+ 1 )
Explanation:
When a sound wave enters a pipe, a resonance process occurs, whereby only some wavelengths can occur.
* in a tube with both ends open
in this case there are maximums at each end, so if the length of the had is L
λ = 2L 1st harmonic
λ = 2 L / 2 2nd harmonic
λ = 2L / 3
λ = 2L / n n = 1, 2, 3,…
* In the case of a tube with one end open and the other closed.
At the open end there is a belly and at the closed end a node
λ = 4L 1st harmonic
λ = 4L / 3 3rd harmonic
λ = 4L / 5
λ = 4L / n n = 1, 3, 5,… odd
n = (2n ’+ 1 ) where n’ are all integers
