A 50.0 kg child stands at the rim of a merry-go-round of radius 1.50 m, rotating with an angular speed of 3.00 rad/s. (a) what i
1 answer:
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The free-body diagram of the forces acting on the flag is in the picture in attachment.
We have: the weight, downward, with magnitude
the force of the wind F, acting horizontally, with intensity
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
By dividing the second equation by the first one, we get
From which we find
which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope:
We have: the weight, downward, with magnitude
the force of the wind F, acting horizontally, with intensity
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
By dividing the second equation by the first one, we get
From which we find
which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope:
Answer:
The quantity of electrons that flows past a given point is 3.0 C.
Explanation:
An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).
i.e I =
It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.
Given that; I = 1.5A and t = 2s, find Q.
Q = It
= 1.5 × 2
= 3.0 C
The quantity of electrons that flows past a given point is 3.0 C.