Answer:
a) the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
Explanation:
Given that;
Gravitational acceleration g = 9.81 m/s²
Mass m = 5 kg
Extension of the spring X = 50 mm = 0.05 m
Spring constant k = ?
we know that;
mg = kX
5 × 9.81 = k(0.05)
k = 981 N/m
a)
Given that; Acceleration of the elevator a = 2 m/s² upwards
Extension of the spring in this situation = X1
Force exerted by the spring = F
we know that;
ma = F - mg
ma = kX1 - mg
we substitute
5 × 2 = 981 × X1 - (5 ×9.81 )
X1 = 0.06019 m
X1 = 60.19 mm
Therefore the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
B)
Acceleration of the elevator = a
The spring is relaxed i.e, it is not exerting any force on the box.
Only the weight force of the box is exerted on the box.
ma = mg
a = g
a = 9.81 m/s² downwards.
Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
Answer:
decantation, distilling, freezing
That measure is known as "acceleration"
a = ΔV / t
Hope this helps!
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 = 
ω2 = 216.47
Answer:
<em>The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.</em>
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.
