The real world application of the atwood’s machine is to be
able to demonstrate specific principles such as the acceleration and dynamics
in a way to understand and know how it is being demonstrated in the real world
and in different situations.
<span>Work done on charge is W = Eqd = σ/(2ε₀) x q x d = {(8.00 x 10⁻¹²)/(2 x 8.854187 x 10⁻¹²)} x 3.00 x 10⁻⁶ x (0.650 - 0.250) = 5.42116402J. KE of sphere = 0.5mv² = 0.5 x 5.00 x 10⁻⁷v² = work done by E-field on charge during its fall = 5.42116402→ v = 4657 m/s.</span>
Answer:
F > W * sin(α)
Explanation:
The force needed for the box to start sliding up depends on the incline (α).
The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.
These forces can be decomposed on their normal and tangential (to the slide plane) components.
The weight will be split into
Wn = W * cos(α) (in normal direction)
Wt = W * sin(α) (in tangential direction)
The normal reaction will be alligned with the normal axis, and will be equal to -Wn
N = -W* cos(α) (in normal direction)
To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger
F > |-W * sin(α)| (in tangential direction)
Answer: set up a reminder so you won't forget also try clearing your schedule so you can have more time
Answer:
The answer is D.
Explanation:
This is because mass always remains constant and weight is dependent on a gravitational pull.
