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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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Answer:

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The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

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= 0.750π(Dd)²/16ρ

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