The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0
y=44.1*100/2 = 2205m
hence, the speed will be
v=0 + a*t = 441m/s
from that height it will just be subjected to the gravitational acceleration
0=v_acc^2 -2g*y_free
y_free = v_acc^2/2g = 9922.5m
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
Answer:
Explanation:
Let the initial velocity of small block be v .
by applying conservation of momentum we can find velocity of common mass
25 v = 75 V , V is velocity of common mass after collision.
V = v / 3
For reaching the height we shall apply conservation of mechanical energy
1/2 m v² = mgh
1/2 x 75 x V² = 75 x g x 10
V² = 2g x 10
v² / 9 = 2 x 9.8 x 10
v² = 9 x 2 x 9.8 x 10
v = 42 m /s
small block must have velocity of 42 m /s .
Impulse by small block on large block
= change in momentum of large block
= 75 x V
= 75 x 42 / 3
= 1050 Ns.
Answer:
Explanation:
Path difference for the observer = 2.04μm
For constructive interference
path difference = n λ
2.04μm = 5 x .408 or , 4 x .51 or 3 x .68μm
For visible wavelengths like .408 , .51 or .68 μm (400 to 700 nm) will the observer see the brightest light due to constructive interference.
B) If the two sources were not in line with the observer, but were still arranged so that one source is 2.04μm farther away from the observer than the other , in that case also , interference pattern will be visible .
c )
For destructive interference
2.04 μm = (2n+1)λ / 2
2.04μm = 3.5 x .58μm
2.04 μm = 4.5 x .45μm
So for .58μm and .45μm , destructive interference will take place
If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.
