<h2>
Answer: 2 lights (in parallel)</h2>
Explanation:
When light bulbs are connected in series the total voltage will be:
(1)
This means the <u>total voltage will be distributed among each of the bulbs</u> in series, therefore the luminosity will be also divided among the bulbs.
When lights are connected in parallel the total voltage will be:
(2)
This means the <u>total voltage will be the same for each bulbs</u> in parallel, therefore the luminosity will be the same.
Now, if we combine this two types of connections, as in this problem with 11 lights in series, which are also in series with 2 lights in parallel; the 2 lights in parallel will be brighter than the 11 lights in series.
<span>The friction force ( µ . N ) must be enough to counter the gravity force ( m . g )
The Normal force from the rotating wall is the centripetal force = mv^2 / r
so with the lowest coefficient of friction ( 0.4 )
0.4 . v^2 / 2.5 = g
v^2 = 2.5 . 9.8 / 0.4
v = 7.83 m/s
T = 5Ď€ / 7.83 = 2.007 s
f = 0.5 Hz = 30 rpm</span>
Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s