Answer:
The induced emf in the loop is 
Explanation:
Given that,
Length of the wire, L = 1.22 m
It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L
4a = 1.22
a = 0.305 m
Area of square, 
Circumference of the loop,

Area of circle,

The induced emf is given by :

So, the induced emf in the loop is 
Explanation:
The changes can be made in airplane longitudinal control to maintain altitude while the airspeed is being decreased is
We can increase the angle of attack this would compensate for the decreasing lift. As the angle of attack directly controls the distribution of pressure on the wings. Moreover, increase in angle of attack will also cause the drag to increase.
Answer:
I will choose the answer A
According to law of conservation of mass within a reaction,
The mass of the compound formed is (23+35.5) grams means 58.5 grams of sodium chloride[NaCl] will be formed.
Answer:
Part a)

Part b)

Explanation:
As we know that torque is defined as the product of force and its perpendicular distance from reference point
so here we have

now we have


Part b)
Now we know the conversion as
1 meter = 3.28 foot
1 N = 0.225 Lb force
now we have


