Answer:
The light used has a wavelenght of 4.51×10^-7 m.
Explanation:
let:
n be the order fringe
Ф be the angle that the light makes
d is the slit spacing of the grating
λ be the wavelength of the light
then, by Bragg's law:
n×λ = d×sin(Ф)
λ = d×sin(Ф)/n
λ = (3.2×10^-4 cm)×sin(25.0°)/3
= 4.51×10^-5 cm
≈ 4.51×10^-7 m
Therefore, the light used has a wavelenght of 4.51×10^-7 m.
Answer:
<em><u>172,000 second </u></em>
<em><u>I'M</u></em><em><u> </u></em><em><u>NOT</u></em><em><u> </u></em><em><u>SURE</u></em><em><u> </u></em><em><u>THAT</u></em><em><u> </u></em><em><u>THIS</u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>RIGHT</u></em><em><u> </u></em><em><u>OR</u></em><em><u> </u></em><em><u>WRONG</u></em><em><u> </u></em><em><u> </u></em><em><u>IF</u></em><em><u> </u></em><em><u>IT'S</u></em><em><u> </u></em><em><u>WRONG</u></em><em><u> </u></em><em><u>THEN</u></em><em><u> </u></em><em><u>SORRY</u></em><em><u> </u></em>
Answer:
Torque decreases .
Explanation:
The tape is pulled at constant speed , speed v is constant , so there is
v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .
Let it be α . Let I be moment of inertia of reel .
Reel is in the form of disc
I = 1/2 m r²
I x α = torque
1/2 m r² x α = torque
As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .
Answer: 580 N
Refer to attached figure.
The angle of inclination is 22 degrees
weight (gravitational force) acts downwards.
Normal force is a contact force which acts perpendicular to the point of contact.
The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.
Gravitational force on an object = mg
The normal force 

Answer:
<em>1108.464 N of force</em>
Explanation:
diameter of water hose = 70 cm = 0.7 m
radius = 0.7/2 = 0.35 m
volumetric flow rate Q = 420 L/min
1 L = 0.001 m^3
1 min = 60 s
therefore,
Q = 420 L/min = (420 x 0.001)/60 = 0.007 m^3/s
Area A of fire hose = π
= 3.142 x
= 0.38 m^2
<em>From continuity equation, Q = AV</em>
where V1 is the velocity of the water through the pipe, and A1 is the area of the pipe.
Q = A1V1
0.007 = 0.38V1
V1 = 0.007/0.38 = 0.018 m/s.
Nozzle diameter = 0.75 cm = 0.0075 m
radius = 0.00375
Area = π
= 3.142 x
= 4.42 x
m^2
velocity of water through the nozzle will be
V2 = Q/A2 = 0.007 ÷ (4.42 x
) = 158.37 m/s
From
<em>F = ρQ(v2 - v1)</em>
Where,
F = force exerted
p = density of water = 1000 kg/m^3
F = 1000 x 0.007 x (158.37 - 0.018) = <em>1108.464 N of force</em>