A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60
1 answer:
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At x = 1.72 units the net force on a charge Q equal to zero when a charge +3q is placed at x = 0 and a charge +11q is placed at x = 5 units.
A net force is defined as the sum of all the forces acting on an object. The equation below is the sum of N forces acting on an object. There may be several forces acting on an object, and when you add up all of those forces, the result is what we call the net force acting on the object.
This equation is the sum of n forces acting on an object. The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula.
+3q ⇔ +11q
x = ? ⇔ x = 5
∴ At x = 1.72 units the net force on a charge Q equal to zero
Learn more about net force here
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Answer:
v = 17.30 m / s
Explanation:
For this exercise we will use Newton's second law
at the bottom of the loop and stopped
∑ F = 0
N-W = 0
N = W
W = 770 N
the mass of the body is
W = mg
m = W / g
m = 770 / 9.8
m = 78.6 kg
on top of the loop and moving
∑ F = m a
N + W = m a
note that the three vectors go in the same vertical direction down
the centripetal acceleration is
a = v² / r
we substitute
N + W = m v² / r
v =
let's calculate
v =
v = 17.30 m / s