Answer: If x + y = a, xxy = b and x • a = 1 , then 2 (a~ - l)a- a x b (b2 ... xy-plane, then the vector in the same plane having projections
Derived quantities depend on.( fundamental)..........physical quantity
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It doesn't because when u threw it the first time, u notice that the ball eventually came to a stop because of the force that was acting upon it. Although when u throw it harder it will start out faster than the first time u threw it because u put more kinetic energy onto the ball. But the same thing happens with this ball that happened to the second ball, they both have a type of force acting upon them.
Answer:
<em>Time period of pendulum is 2.02 s.</em>
Explanation:
A <em>simple pendulum</em> is a device which consists of mass m hanging from the string of length L attached to the some point.When displaced and released its swings back and forth with periodic motion.
The time period of pendulum is defined as time taken by the pendulum to complete one full oscillation . it is denoted by T.
By <em>Huygens law of period of pendulum</em>,
T = 2π
eqn 1
where L is the length of pendulum,
g is acceleration due to gravity
<em>Period of pendulum is independent of the mass of pendulum,</em>
<em />
Substituting values in eqn 1
T = 2π 
T = 2.02 s
<em>Time period of pendulum is 2.02 s.</em>
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
