<span>An object roating at one revokution per second has an angular velocity of 360 degrees per second or 2pi radians per second. This is found by taking the number of revolutions over a period of time and than dividing by the chosen period of time to get the velocity. There are 360 degrees or 2pi radians in one revolution.</span>
Answer:
The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.
Explanation:
The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.
This is because:
If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.
Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.
Therefore, static friction is infact necessary for a ball to begin rolling.
Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.
So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.
Answer:
a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈
Explanation:
For this exercise we use ohm's law
V = i R
i = V / R
we assume that the applied voltage is the same in all cases
let's find the current for each resistance
R = 4 Ω
i₄ = V / 4
R = 8 Ω
i₈ = V / 8
we look for the relationship between these two currents
i₈ /i₄ = 4/8 = ½
i₈ = 0.5 i₄
R = 3 Ω
i₃ = V3
R = 10 Ω
i₁₀ = V / 10
we look for relationships
i₁₀ / 1₃ = 3/10
i₁₀ = 0.3 i₃
i₁₀ / 1₈ = 8/10
i₁₀ = 0.8 i₈
The resistance of a single light bulb is 220 ohms per bulb.
<h3>What is Ohm's Law?</h3>
Ohm's Law is a formula used to determine how voltage, current, and resistance in an electrical circuit relate to one another.
Ohm's Law (E = IR) is as basic to students of electronics as Einstein's Relativity equation (E = mc2) is to physicists.
E = I x R
The formula reads voltage = current x resistance, or V = A xΩ., or volts = amps x ohms.
110volts divided by .25amps = 440 ohms. 440 divided by 2 =220 ohms per bulb.
R = 110/(2*0.25) = 220 ohms
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Answer:
a. Zin = 41.25 - j 16.35 Ω
b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c. Pin = 216 w
d. PL = Pin = 216 w
e. Pg = 478.4 w , Pzg = 262.4 w
Explanation:
a.
Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]
βl = 2π / λ * 0.15 λ = 54 °
Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]
Zin = 41.25 - j 16.35 Ω
b.
I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶
V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)
V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c.
Pin = ¹/₂ * Re * [V₁ * I₁]
Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)
Pin = 216 w
d.
The power PL and Pin are the same as the line is lossless input to the line ends up in the load so
PL = Pin
PL = 216 w
e.
Pg Generator
Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)
Pg = 478.4 w
Pzg dissipated
Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50
Pzg = 262.4 w