Answer:
W ≅ 292.97 J
Explanation:
1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)
Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;
W = (Fcosθ)d
Given that:
Tension of the force = 62 N
angle of incline θ = 34°
distance d =5.7 m.
Then;
W = 62 × cos(34) × 5.7
W = 353.4 cos(34)
W = 353.4 × 0.8290
W = 292.9686 J
W ≅ 292.97 J
Hence, the work done by tension before the block goes up the incline = 292.97 J
Answer:
JA
Explanation:
s of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position. Assume all variable and constants are in SI units.
Answer:
0.050 m
Explanation:
The strength of the magnetic field produced by a current-carrying wire is given by
where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.
In this problem, we have:
(current in the wire)
(strength of magnetic field)
Solving for r, we find the distance from the wire:
