Answer:
= 625 nm
Explanation:
We now that for
for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....
d= distance between the slits, λ= wavelength of incident ray
for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.
Given
d=1.19 mm, y=4.97 cm, and, n=10, D=9.47 m
applying formula
λ= (d*y)/(D*n)
putting values we get
on solving we get
= 625 nm
Answer:
False
Explanation:
No. The buoyant force on an object is the portion of its weight that appears to vanish
when the object is in any fluid (could be either a liquid or a gas).
If the object happens to float in a particular fluid, then the buoyant force at that moment
is equal to the object's weight.
Notice that the buoyant force on an object will be different in different fluids.
Answer: 4100 Mpc
Explanation:
Since H o = 70 km/s/Mpc
Redshift z = 5.82
Recessional velocity vr = 287,000 km/s
Then, the distance to the galaxy in light years will be:
= Recessional velocity / H o
= 287000 / 70
= 4100 Mpc
Answer:
A = m³/s³ = [L]³/[T]³ = [L³T⁻³]
B = m³s = [L³T]
Explanation:
We have the equation:
V = At³ + B/t
where, the dimensions of each variable are as follows:
V = m³ = [L]³
t = s = [T]
substituting these in equation, we get:
m³ = A(s)³ + B/s
for the homogeneity of the equation:
A(s)³ = m³
<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>
Also,
B/s = m³
<u>B = m³s = [L³T]</u>
