Drag a battery into the construction panel, and use the voltmeter to determine which end of the battery is the positive terminal
1 answer:
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Answer:
a = 3.27 m/s²
F = 32.7 N
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling straight down.
Normal force N pushing perpendicular to the slope.
Friction force F pushing parallel up the slope.
Sum of forces in the parallel direction:
∑F = ma
mg sin θ − F = ma
Sum of torques about the cylinder's axis:
∑τ = Iα
Fr = ½ mr²α
F = ½ mrα
Since the cylinder rolls without slipping, a = αr. Substituting:
F = ½ ma
Two equations, two unknowns (a and F). Substituting the second equation into the first:
mg sin θ − ½ ma = ma
Multiply both sides by 2/m:
2g sin θ − a = 2a
Solve for a:
2g sin θ = 3a
a = ⅔ g sin θ
a = ⅔ (9.8 m/s²) (sin 30°)
a = 3.27 m/s²
Solving for F:
F = ½ ma
F = ½ (20 kg) (3.27 m/s²)
F = 32.7 N
COMPLETE QUESTION:
<em>When the magnetic flux through a single loop of wire increases by </em>30 Tm^2<em> , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of </em><em>2.5 ohms </em><em>, (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?</em>
Answer:
(a). The time period is 0.3s.
(b). The time period is 0.6s.
Explanation:
Faraday's law says that for one loop of wire the emf is
and since from Ohm's law
,
then equation (1) becomes
(a).
We are told that the change in magnetic flux is , the current induced is
, and the resistance of the wire is
; therefore, equation (2) gives
which we solve for to get:
,
which is the period of time over which the magnetic flux increased.
(b).
Now, if the current had been , then equation (2) would give
which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux is greater than in part b, and therefore , the current in (a) is greater than in (b).