The main cause of this is Friction. The more oil that is laid down, the less friction there is between the ball and the lane surface. The less friction, the harder it is for the bowler to send the ball in a curved path imparted by the spin that the bowler puts on the ball at the instant of release.
-Synodic period is the period of celestial bodies observed on the moving planet(mostly earth)
Sideral period is the period comparing to the fixed stars without motion of the earth involved.
(I will explain the second question with an example, so it's easier to understand)
-For Sideral month for example of the moon it cactually complete one revolution in around 27.3 days.
However, since the earth moves, for us it took some more time to see the moon the same as before (fullmoon to fullmoon) again. That make synodic month of the moon to be around 29.5 days.
The answer is A. A folkway is a closely held belief by a specific group of people.
<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>
When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:

To maintain
constant, if Q increases V also increases.
On the other hand, the potential energy
can be expressed as:

In conclusion, as Q increases the potential energy also increases.
Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement

putting values h=15 m, v=0.8
............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get

solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = 
⇒ 
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m