See projectiles are very simple unless you understand its core concepts....projectile is nothing just mixture of upward motion and horizontal motion....
THE KEY IS FORGET THE NAME PROJECTILE...ITS JUST HORIZONTAL MOTION + VERTICAL MOTION
Answer:
The angular acceleration of the pencil<em> α = 17 rad·s⁻²</em>
Explanation:
Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:
τ = I α (1)
W r = I α (2)
The weight is that the pencil has is,
sin 10 = r / (L/2)
r = L/2(sin(10))
The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:
I = 1/3 M L²
Thus,
mg(L / 2)sin(10) = (1/3 m L²)(α)
α(f) = 3/2(g) / Lsin(10)
α = 3/2(9.8) / 0.150sin(10)
<em> α = 17 rad·s⁻²</em>
Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>
Answer:
$ 0.48
Explanation:
We can calculate this quantity easily using successive products and taking into account the units.
![\frac{0.08}{kw*h}*2[kw]*3[hr]\\ \\=0.48](https://tex.z-dn.net/?f=%5Cfrac%7B0.08%7D%7Bkw%2Ah%7D%2A2%5Bkw%5D%2A3%5Bhr%5D%5C%5C%20%5C%5C%3D0.48)
The amount is $ 0.48
Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop, IR = E- V
IR = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:
IR = V
Therefore, the resistance that will provide this potential drop is 388.89 ohms.
1 pound ≈ 0.4536 kg
170 pounds ≈ 170 * 0.4536 kg
≈ 77.112 kg
