We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
A scientific theory is different from a hypothesis because a theory is a educated guess that is being worked on and proven correct and a hypothesis is a educated guess it is a guess that needs to be proven.
Answer:
Explanation:
= Torque = 36.5 Nm
= Initial angular velocity = 0
= Final angular velocity = 10.3 rad/s
t = Time = 6.1 s
I = Moment of inertia
From the kinematic equations of linear motion we have
Torque is given by
The wheel's moment of inertia is
t = 60.6 s
= 10.3 rad/s
= 0
Frictional torque is given by
The magnitude of the torque caused by friction is
Speeding up
Slowing down
Total number of revolutions
The total number of revolutions the wheel goes through is .
Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy 
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
The answer would be D. Because the iron fillings are attracted to the magnet underneath the glass.
