Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm
Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as
y = R×(mλ/d)
Where y = distance between nth fringes and the center fringe.
m = order of fringe
λ = wavelength of light = 589nm = 589×10^-9m
R = distance between slits and screen = 1.0m
d = distance between slits = 0.25mm = 0.00025m
For distance between the first dark fringe and the center fringe.
This implies that m = 1
y = 1 × 589×10^-9 × 1/0.00025
y = 589×10^-9/0.00025
y = 2,356,000 × 10^-9
y = 2.356 × 10^-3 m = 2.356mm
For the second dark fringe, this implies that m = 2
y = 1 × 2 × 589×10^-9/0.00025
y = 1178 × 10^-9 /0.00025
y = 4,712,000 × 10^-9
y = 4.712 × 10^-3 m = 4.712mm
Answer:
Uniform rectilinear movement (m. r. u.)
Explanation:
It is a continuous movement without acceleration, that is, it moves at a constant speed. The speed does not change over time, for this reason, there is no change in acceleration.
Vf = final velocity = 50/30 [km/s] = 1.67 [km/s]
Vo = initial velocity = 50/30 [km/s] = 1.67 [km/s]
Answer:
c. 0.816
Explanation:
Let the mass of car be 'm' and coefficient of static friction be 'μ'.
Given:
Speed of the car (v) = 40.0 m/s
Radius of the curve (R) = 200 m
As the car is making a circular turn, the force acting on it is centripetal force which is given as:
Centripetal force is, 
The frictional force is given as:
Friction = Normal force × Coefficient of static friction
As there is no vertical motion, therefore, 
. So,
Now, the centripetal force is provided by the frictional force. Therefore,
Frictional force = Centripetal force
Plug in the given values and solve for 'μ'. This gives,
Therefore, option (c) is correct.
Answer:
If the buoyant force is greater than the object's weight, the object will rise to the surface and float. If the buoyant force is less than the object's weight, the object will sink. If the buoyant force equals the object's weight, the object will remain suspended at that depth.
Explanation:
Not much explaining to do here!
