A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour
= Total distance traveled/ total time taken
=
The tension in the string B) It quadruples.
Explanation:
The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:
where:
T is the tension in the string
m is the mass of the ball
v is the speed of the ball
r is the radius of the circle (the lenght of the string)
In this problem, we are told that the speed of the ball is doubled, so
v' = 2v
Substituting into the previous equation, we find the new tension in the string:
Therefore, the tension in the string will quadruple.
Learn more about circular motion:
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Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.