Answer:
width of slit =1.23× 10⁻⁶ m
Explanation:
we know the condition of diffraction minima,
d sin θ = n λ
λ = wavelength θ = angle between the central maxima and 1st minima
d = slit width
for first minima n = 1
now,
d =
d =
d = 1228 × 10⁻⁹ m = 1.228× 10⁻⁶ m
d = 1.23× 10⁻⁶ m
width of slit =1.23× 10⁻⁶ m
Answer:
The Amount of work is done on the box by the net force is 378 Joules.
Explanation:
Given:
A 42 N net force is applied to a box which slides horizontally across a floor.
Force = 42 N
Force applied for a displacement of 9.0 m
Displacement = 9.0 m
To Find:
Amount of work is done on the box by the net force = ?
i.e Work = ?
Solution:
When a force causes a object to move, work is being done on the object by the force. Work is the measure of energy transfer when a force (F) moves an object through a distance (d).
Formula is given by
Substituting the given values we get
∴ Work = 42 × 9.0
∴ Work = 378 N-m
or
∴ Work = 378 Joules
The Amount of work is done on the box by the net force is 378 Joules.
Answer:
The rate of flow of water is 71.28 kg/s
Solution:
As per the question:
Diameter, d = 18.0 cm
Diameter, d' = 9.0 cm
Pressure in larger pipe, P =
Pressure in the smaller pipe, P' =
Now,
To calculate the rate of flow of water:
We know that:
Av = A'v'
where
A = Cross sectional area of larger pipe
A' = Cross sectional area of larger pipe
v = velocity of water in larger pipe
v' = velocity of water in larger pipe
Thus
v' = 4v
Now,
By using Bernoulli's eqn:
where
h = h'
Now, the rate of flow is given by: