Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center
Second minima from center
The distance between the places where the intensity is zero due to the double slit effect
Put the value into the formula
Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Answer:
Part A : E = 
ε₀ Q₁/R₁² Volt/meter
Part B : V = ε₀ Q₁/R₁ Volt
Explanation:
Given that,
Charge distributed on the sphere is Q₁
The radius of sphere is R
₁
The electric potential at infinity is 0
<em>Part A</em>
The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F = ε₀ Q₁/R₁²
Then the electric field at that point is
E = F/1
E = ε₀ Q₁/R₁² Volt/meter
Part B
The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V = ε₀ Q₁/R₁ Volt
Answer:
The group of light rays is reflected back towards the focal point thereby producing a magnifying effect.
Explanation:
The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration
Complete question:
A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?
Answer:
the total units of insulin admistered each morning
= 22 units of qam and humulin
Explanation:
given
44 units and Humnlin N
with concentration 50/100 = 1/2 = 0.5
∴ 44 × 0.5 ≈ 22 units in the morning
regular insulin administered each day
(22 + 35)units of qam and humulin
= 57units
<span>First, we use the kinetic energy equation to create a formula:
Ka = 2Kb
1/2(ma*Va^2) = 2(1/2(mb*Vb^2))
The 1/2 of the right gets cancelled by the 2 left of the bracket so:
1/2(ma*Va^2) = mb*Vb^2 (1)
By the definiton of momentum we can say:
ma*Va = mb*Vb
And with some algebra:
Vb = (ma*Va)/mb (2)
Substituting (2) into (1), we have:
1/2(ma*Va^2) = mb*((ma*Va)/mb)^2
Then:
1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2
We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1:
1/2(ma) = (ma^2)/mb
Cancel the ma of the left, leaving the right one with exponent 1:
1/2 = ma/mb
And finally we have that:
mb/2 = ma
mb = 2ma</span>
