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Lynna [10]
3 years ago
9

A truck with a mass of 1.5 x 103 kg accelerates to a speed of 18.0 m/s in 12.0 s from a dead stop. Assume that the force of resi

stance is a constant 400.0 N during the acceleration. What is the average power developed by the truckÍs engine?
Physics
2 answers:
77julia77 [94]3 years ago
4 0
There are several information's already given in the question. Based on those information's the answer can be easily deduced.
Mass of the truck = <span>1.5 x 10^3 kg
Constant force of resistance = 400 N
Final Velocity = 18 m/s
Initial velocity = 0 m/s
Time taken = 12 s
Then
Acceleration = (Final velocity - initial velocity)/Time
                     = 18 /12
                     = 3/2
                     = 1.5 m/s^2
Then
Driving force = Mass * Acceleration
                     = 1.5 * 10^3 * 1.5
                     = 2250 N
So
Net Force = 2250 - 400
                 = 1850 N
Average power developed = 1850 * 18
                                           = 3.33 * 10^4 W</span>
Yakvenalex [24]3 years ago
3 0
Power = Net Force x velocity
Net force = driving force - force of resistance
Driving force = mass x acceleration
Acceleration = (final velocity - initial velocity) / time
Acceleration = (18 - 0) / 12 = 1.5 m/s²
Driving force = 1.5 x 10³ x 1.5
= 2250 N
Net force = 2250 - 400
= 1850
Power = 1850 x 18
= 3.33 x 10⁴ Watts
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First of all, let's convert the energy of the absorbed photon into Joules:

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In this case, when the electron jumps from the n=4 level to the n=3 level, emits a photon with wavelength

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E=\frac{hc}{\lambda}

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E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J

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a) The initial angular speed is 209.3 m/s

b) The angular acceleration is -1.74 rad/s^2

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

\omega_i = 2000 rpm

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

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We find:

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b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s

Now we can find the angular acceleration, given by

\alpha = \frac{\omega_f - \omega_i}{t}

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\omega_i = 209.3 rad/s is the initial angular speed

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t = 60 s is the time interval

Substituting,

\alpha = \frac{104.7-209.3}{60}=-1.74  rad/s^2

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

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And substituting t = 40 s, we find

\omega' = 209.3 + (-1.74)(40)=139.7 rad/s

d)

The angular displacement of the wheel in a certain time interval t is given by

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\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev

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