In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
Answer:
90 ft/s is what i put. Let me know if its wrong
Answer:
2632 foot-pound
Explanation:
Work done: Work is said to be done when ever a force moves a body through a given distance. The S.I unit of force is Newton (N).
From the question,
The expression for work done is given as,
W = Fdcos∅......................... Equation 1
Where W = work done, F = force, d = distance, ∅ = angle between the force and the horizontal.
Given: F = 32 lbs, d = 90 feet, ∅ = 24°
substitute into equation 1
W = 32×90×cos24
W = 2880(0.914)
W = 2632.32
W = 2632 foot-pound
Answer:
Current needed = 704A
Explanation:
Using the fomula; torque(τ) = (I)(A)(B)Sinθ
Where B = uniform magnetic field
I = current and A = Area
Diameter = 19cm = 0.19m so, radius = 0.19/2 = 0.095m
Area(A) = πr^(2) = πr^(2)
= π(0.095)^(2) = 0.0284 m^(2)
Now, B(earth)= 5x10^-5 T
While, we can ignore the angle because it's insignificant since the angle of the wire is oriented for maximum torque in the earth's field.
Now, if we arrange the formula to solve for charge (I):
I = (τ)/(A)(B)
I = (1.0x10^-3) / (0.0284)(5x10^-5)
I = 704A
