Both of them are magnets coiled by wires.
1. The wire coiled in the first diagram, the wire is having current, Making the magnetic feild of the magnet more........
2. The wire coiling the magnet is here not having electric current making the magnetic feild smaller
31 m/s ÷ 9 m/s² = 3.44 s
Time = Change in velocity divided (÷) by acceleration.
Answer:
A = (27.95 N, 21 N)
Explanation:
The polar co-ordinates are given as:
(r,θ) = (35 N, 37°)
Now, to convert this into polar co-ordinates (x, y), we will use following relations:
r² = x² + y²
(35)² = x² + y²
1225 = x² + y² ----------- equation (1)
and
tan θ = y/x
tan 37° = y/x
y = 0.753 x ------------------- equation (2)
Substituting this value in equation (1):
1225 = x² + (0.753 x)²
1225 = 1.567 x²
x² = 1225/1.567
x = √781.32
x = 27.95 N
using this value in equation (2)
y = (0.753)(27.95 N)
y = 21 N
Therefore, the vector can be represented in polar co-ordinates as:
<u>A = (27.95 N, 21 N)</u>
The kinetic and potential energies must be taken into account. The ball lost some speed but was also raised in the vertical direction between the two measured positions. This will cost some of the kinetic energy that we don't want to ascribe to the air resistance losses.
The gain in potential energy is
mg(h₂-h₁) = 0.583*9.81*(3.10-1.80) = 7.435J
The loss in kinetic energy is
(1/2)m(v₁² - v₂²) = (0.583/2)(7.29²-4.22²)=10.3J
Since 7.435J of this is due to the fact that it gained potential energy the final result is E=10.3J-7.435J=2.865J.
One way to think about the potential energy part is that it is higher in the air at the hoop than when the player releases the ball. In other words it hasn't fallen as far down as it went up, it is still in the process of gaining speed on the way down. This reduction in speed means less kinetic energy, but not because it was lost. It will be gained once it falls to the height of 1.8m, minus the little loss it will experience over that distance due to air resistance.
<h3>
Answer: 28 m/s</h3>
Work Shown:
- Vi = starting or initial velocity = 10 m/s
- Vf = unknown final velocity
- a = acceleration = 3 m/s^2
- t = duration of time = 6 seconds
Use those values in the following kinematic equation
Vf = Vi + a*t
Vf = 10 + 3*6
Vf = 10 + 18
Vf = 28
The final velocity is 28 m/s
