I think jogging/running/walking because you don't need any equipment and you can structure around it on your own time.
Answer: B. The particles mover faster.
Depending on which type of motor you're talking about, but the first 3 are true. A stronger magnetic field in a DC motor will slow it down but increase its torque.
The amount of current in the motor will control the magnetic fields and therefore affect the speed (and torque). In an induction motor, the rotational speed is given by

where f is the line frequency and p is the number of poles. Thus fewer poles makes it go faster.
Answer:
a) 
Explanation:
a) Let assume that the ground is not inclined, since the bottom of the playground slide is tangent to ground. Then, the length of given by the definition of a circular arc:



The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:


The average frictional force is cleared within the expression:

![f = \frac{(12\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)-\frac{1}{2}\cdot (4.5\,\frac{m}{s} )^{2} ]}{6.676\,m}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B%2812%5C%2Ckg%29%5Ccdot%20%5B%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%29%5Ccdot%20%283%5C%2Cm%29-%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%284.5%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%20%5D%7D%7B6.676%5C%2Cm%7D)

Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres