I think it would be 77.9 grams
Answer:
λ = 1.4 × 10^(-7) m
Explanation:
We are given;
distance of eye piece from the source;D = 1.5 m
distance between the virtual sources;d = 7.5 × 10^(-4) m
To find the wavelength, we will use the formula for fringe width;
X = λD/d
Where X is fringe width, λ is wavelength, while d and D remain as before.
Now, fringe width = eye-piece distance moved transversely/number of fringes
Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m
Thus,
Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m
Thus;
1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))
λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5
λ = 1.4 × 10^(-7) m
Answer:
Pressure applied to the needle is 7528 Pa
Explanation:
As we know by poiseuille's law of flow of liquid through a cylindrical pipe
the rate of flow through the pipe is given as

now we know that

radius = 0.2 mm
Length = 6.32 cm

now we have



now we have


Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ =
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m