I and II only it’s has multiple paths for the electricity to flow
I believe it's the the third one. :)
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Answer:
force (tension) of 29.4 N (upward) in 100 cm
force (tension) of 58.4 N (upward) in 200 cm
Explanation:
Given:
Length of tube = 5 m (500 cm)
Mass of tube = 9
Suspended vertically from 150 cm and 50 cm.
Computation:
Force = Mass × gravity acceleration.
Force = 9.8 x 9
Force = 88.2 N
So,
Upward forces = Downward forces
D1 = 150 - 50 = 100 cm
D2 = 150 + 50 = 200 cm
And F1 = F2
F1 x D1 = F2 x D2
F1 x 100 = F2 x 200
F = 2F
Total force = Upward forces + Downward forces
3F = 88.2
F = 29.4 and 2F = 58.8 N
force (tension) of 29.4 N (upward) in 100 cm
force (tension) of 58.4 N (upward) in 200 cm
Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
