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kumpel [21]
3 years ago
6

Problem 16.71 A bowler projects an 8-in.-diameter ball weighing 12 lb along an alley with a forward velocity v0 of 15 ft/s and a

backspin 0 of 9 rad/s. Knowing that the coefficient of kinetic friction between the ball and the alley is 0.10, determine (a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the ball at time t1, (c) the distance the ball will have traveled at time t1.

Physics
1 answer:
alex41 [277]3 years ago
4 0

Answer:

Find the attachments for complete solution

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I think it means the object is slowing down
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Two long, straight wires are separated by 0.12 m. The wires carry currents of 4.0 A in opposite directions, as the drawing indic
user100 [1]
If we have I= 7.5 A:

I think my solution might just help you answer the problem on your own:

You have the formulas correct, watch your signs and BRACKETS. 

B = μ0/(2π) (Current) / (Perpendicular distance) 
Since μ0=4π E -7 Tm/A, we have: 
B1 = (4πE-7 Tm/A)(7.5 A)/[2π (0.030 m)] = 5E-5 T 
B2 = (4πE-7 Tm/A)(-7.5 A)/[2π (0.150 m)] = -1E-1 T 
So BA = B1 + B2 = ? 
(It looks like you just left out the square brackets, hence multiplying Pi and 0.03 and 0.15 instead of dividing them.) 

<span>For the point B, the two distances are -0.060 m and +0.060 m. Be careful with the signs. Unlike point A, the two components will have the same sign.</span>
3 0
3 years ago
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Some of the energy the car gained as it was pulled was lost because of heat from friction. The rest of the energy was transforme
Lisa [10]

Answer:

454,320 joules

Explanation:

The work done on an object is equal to its change in kinetic energy: Change in KE = F × d.

Plug the values for F and d into the formula and solve:

Change in KE = 2,524 × 180

= 454,320 joules

The roller coaster gains 454,320 joules of energy from the work done on it by the chain.

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4 years ago
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If the current decreases uniformly from 5.00 a to 2.00 a in 3.00 ms, calculate the self-induced emf in the coil.
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<span>If the current decreases uniformly from 5.00A to 2.00 A in 3.00ms , the self-induced emf in the coil. =70.02025 V    </span>
5 0
3 years ago
An AC source operating at 59 Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and an induc
Alexxandr [17]

I = V/Z

V = voltage, I = current, Z = impedance

First let's find the total impedance of the circuit.

The impedance of the resistor is:

Z_{R} = R

R = resistance

Given values:

R = 1200Ω

Plug in:

Z_{R} = 1200Ω

The impedance of the inductor is:

Z_{L} = j2πfL

f = source frequency, L = inductance

Given values:

f = 59Hz, L = 2.4H

Plug in:

Z_{L} = j2π(59)(2.4) = j889.7Ω

Add up the individual impedances to get the Z, and convert Z to polar form:

Z = Z_{R} + Z_{L}

Z = 1200 + j889.7

Z = 1494∠36.55°Ω

I = V/Z

Given values:

V = 170∠0°V (assume 0 initial phase)

Z = 1494∠36.55°Ω

I = 170∠0°/1494∠36.55°Ω

I = 0.1138∠-36.55°A

Round the magnitude of I to 2 significant figures and now you have your maximum current:

I = 0.11A

5 0
3 years ago
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