Answer:
The force constant of the spring is 317.8 N/m.
Explanation:
Given that,
Frequency 
We need to calculate the reduced mass
Using formula of reduced mass
Where,
= atomic mass of H
= atomic mass of I
Put the value into the formula




We need to calculate the force constant of the spring
Using formula of frequency


Put the value into the formula


Hence, The force constant of the spring is 317.8 N/m.
Answer:
The magnitude of the velocity of the aircraft P relative to aircraft Q is zero
Explanation:
The velocity of the two aircraft, P & Q, v = 300 m/s
The angle of the direction between them, Ф = 90°
The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula
<em> V = v cos Ф
</em>
Substituting the values in the above equation
v = 300 x cos 90°
= 300 x 0
= 0
Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.
Answer:
When she adds more washers to the meter, the magnitude of force that is shown on the force meter increases.
Explanation:
The force that the washers exert on the force meter is actually the weight of the washers. Weight is actually a force with gravitation acceleration.
F = W = mg
Where g is gravitational acceleration and its value is 9.81 m/s² and m is the mass of any object. As she adds more washers to the meter so the total mass of the washers increases. As the mass of the washers increases, magnitude of the force (Weight) shown on the force meter increases.
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
Answer:
Part a)

Part b)

Explanation:
Part a)
Electric field due to large sheet is given as







now the electric field is given as


Part b)
Now since the electric field is required at same distance on other side
so the field will remain same on other side of the plate
