In a particular machine, there are 2 gears that interlock; One gear is larger in circumference than the other. The manufacturer
1 answer:
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In a particular machine, there are 2 gears that interlock; One gear is larger in circumference than the other. The manufacturer of the gears guarantees that each gear will last for at least 6,000,000,000 revolutions. Assuming that there is no slippage between the 2 gears and that when one gear rotates the other gear also rotates, the larger gear is guaranteed to last how many days longer than the smaller gear?
(1) The diameter of the larger gear is twice the diameter of smaller gear.
(2) The smaller gear revolves 600 times per minute.
Answer:
Number of days larger gear last longer than small gear=115.74 days
Explanation:
Given Data
Revolution=6,000,000,000 revolutions
diameter of larger gear is twice diameter of smaller gear
Smaller gear revolves=600 times per minute
Number of days larger gear last longer than small gear=?
Solution
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(a)
The resistance of the rod is given by:
(1)
where
is the material resistivity
L = 1.20 m is the length of the rod
A is the cross-sectional area
The radius of the rod is half the diameter: , so the cross-sectional area is
The resistance at 20°C can be found by using Ohm's law. In fact, we know:
- The voltage at this temperature is V = 15.0 V
- The current at this temperature is I = 18.6 A
So, the resistance is
And now we can re-arrange the eq.(1) to solve for the resistivity:
(b)
First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is
I = 17.5 A
So the resistance is
The equation that gives the change in resistance as a function of the temperature is
where
is the resistance at the new temperature (92.0°C)
is the resistance at the original temperature (20.0°C)
is the temperature coefficient of resistivity
Solving the formula for , we find
Answer:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e
N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e
= 1250W
d) Final peak=
P= Ik/e
=
P = 2.5 × 10⁷W