The answer would be, the iris. Hope this helps:)
Answer:
R = 7 [amp]
Explanation:
To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance. In this way, we have the following equation.
V = I*R
where:
V = voltage = 49 [V] (units of volts)
I = current = 7 [amp] (amperes)
R = resistance [ohms]
Now clearing R.
R =V/I
R = 49/7
R = 7 [amp]
letter a because 4N to the left +3N to the right = 4N left minus 3N left as you change the direction of 3N then you also change the plus into minus. the direction will always favour to which of the direction is bigger
This first step involves a right triangle. If the plane is 400 km east and 300 km south of the origin, and it flew in a straight line, then you can construct a right triangle with side lengths 300, 400, and c. You may recognize that these are multiples of the Pythagorean triple 3, 4, 5, so the side length c is 500 km. Otherwise, you would write

.
This second step is, if I am correctly interpreting "degrees south of east," to find the angle formed by the horizontal line representing the east and the path of the plane. I made a diagram that does just that (see attached). You can use a trig function of one of the angles to solve. I chose

. Thus, I believe it is 37° south of east.
Answer:
72.75 kg m^2
Explanation:
initial angular velocity, ω = 35 rpm
final angular velocity, ω' = 19 rpm
mass of child, m = 15.5 kg
distance from the centre, d = 1.55 m
Let the moment of inertia of the merry go round is I.
Use the concept of conservation of angular momentum
I ω = I' ω'
where I' be the moment of inertia of merry go round and child
I x 35 = ( I + md^2) ω'
I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19
35 I = 19 I + 1164
16 I = 1164
I = 72.75 kg m^2
Thus, the moment of inertia of the merry go round is 72.75 kg m^2.