A uniform marble rolls down a symmetric bowl, starting from rest atthe top of the left side. The top of each side is a distanceh
above the bottom of the bowl. The left half of the bowlis rough enough to cause the marble to roll without slipping, butthe right half has no friction because it is coated with oil.
(a) How far up the smooth side will the marblego, measured vertically from the bottom? (Use the followingvariable as necessary: h.)
H = 1Click here to preview your answer.
(b) How high would the marble go if both sides were as rough as theleft side? (Use the following variable as necessary: h.)
H = 2Click here to preview your answer.
(c) How do you account for the fact that the marble goeshigher with friction on the right side than withoutfriction?
(a) How far up the smooth side will the marblego, measured vertically from the bottom? (Use the followingvariable as necessary: h.)
H = 1Click here to preview your answer.
(b) How high would the marble go if both sides were as rough as theleft side? (Use the following variable as necessary: h.)
H = 2Click here to preview your answer.
(c) How do you account for the fact that the marble goeshigher with friction on the right side than withoutfriction?
1 answer:
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1. The voltage on the circuit with a resistance of 12.5 Ω and current of 2.35 A is 29.4 V.
2.The resistance in the circuit is found to be 1.45 Ω.
3. The equivalent resistance of the resistors connected in series is 24 Ω.
4. The equivalent resistance of the resistors connected in parallel is 2.18Ω.
5. The power generated is 552.5 W.
6. The frequency of the green light is 0.56×10¹⁵ Hz.
Explanation:
1) This problem can be solved using Ohm's law. Here the resistance (R) of the circuit is given to be 12.5 Ω and the current (I) is stated to be 2.35 A. So the ohm's law states that in a closed circuit, the voltage will be directly proportional to the current flowing in the circuit and the resistance will act as the proportionality constant.
So, the voltage on the circuit with a resistance of 12.5 Ω and current of 2.35 A is 29.4 V.
2) Using the same Ohms' law, now we have to determine the resistance. So in this case, the voltage is given as 9 V and the current is said to be 6.2 A, then resistance can be determined as the ratio of voltage to current.
So, the resistance in the circuit is found to be 1.45 Ω.
3) Here, the resistances of three resistors are given as 4 Ω, 8 Ω and 12 Ω. And it is stated that the resistances are connected or wired in series. Then the equivalent resistance will be obtained by the sum of resistances of three resistors, as the current flow will be constant in all the three resistors.
Thus, the equivalent resistance of the resistors connected in series is 24 Ω.
4) Now, if the resistors are connected in parallel, then the equivalent resistance will be ratio of product of resistances to the sum of the resistances.
Thus, the equivalent resistance of the resistors connected in parallel is 2.18Ω.
5) Power generated by the person can be obtained by the ratio of work done by the person to the time in which the work is done. So the work done can be obtained by the product of force with displacement.
As here the weight lifted by the person will act as dominant force on the person. So the force is considered as F = 956 N and the displacement is d = 2.41 m, then
So, the work done is obtained as 2303.96 J and the time is given as 4.17 s, then
So, the power generated is 552.5 W.
6) In this, the wavelength of green light is given as 5.34 × 10⁻⁷ m. It is known that the wavelength is inversely proportional to the frequency.
As, speed of light is known as 3×10⁸ m/s, the frequency will be determined as the ratio of speed of light to wavelength.
Thus, the frequency of the green light is 0.56×10¹⁵ Hz.