Answer:
3 : 08 : 10.9
Explanation:
assuming a 12 hour clock
angular velocity of the hour hand is 2π/(3600(12)) rad/s
angular velocity of the minute hand is 2π/3600 rad/s
difference is 2π/3600 - 2π/(3600(12)) = 11(2π/(3600(12)) rad/s
45° = π/4 radians
This angle is covered in a time of
π/4 rad / 22π / (3600(12)) = 900(12) / 22 = 490.909090... s
or about 8 minutes 10.9 s
ANSWER 3:08:10.9
k = 5.29
a = 0.78m/s²
KE = 0.0765J
<u>Explanation:</u>
Given-
Mass of air tracker, m = 1.15kg
Force, F = 0.9N
distance, x = 0.17m
(a) Effective spring constant, k = ?
Force = kx
0.9 = k X0.17
k = 5.29
(b) Maximum acceleration, m = ?
We know,
Force = ma
0.9N = 1.15 X a
a = 0.78 m/s²
c) kinetic energy, KE of the glider at x = 0.00 m.
The work done as the glider was moved = Average force * distance
This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0
As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)
Work = Kinetic energy
KE = 0.450 * 0.17
KE = 0.0765J
Answer: 477W
Explanation:
Given the following :
Mass (m) = 7.3kg
Initial Velocity (u) = 0
Final velocity (v) = 14m/s
time (t) = 1.5s
Power = workdone (W) / time (t)
The workdone can be calculated as the change in kinetic energy (KE) :
Recall ;
KE = 0.5mv^2
Therefore, change in KE is given by:
0.5mv^2 - 0.5mu^2
Change in KE = 0.5(7.3)(14^2) - 0.5(7.3)(0^2)
Change in KE = 715.4J
Therefore ;
Average power = Workdone / time
Workdone = change in KE = 715.4N
Average power = 715.4 / 1.5
Average power = 476.93333 W
= 477W
So basically you, then, finally, you
