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3 years ago

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A car with a mass of 1240kg starts from rest and accelerates to a speed of 19.4m/s in 10.2s. Assume that the force of resistance

remains constant at 376N during this time. What is the average power developed by the car's engine?

1 answer:

garri49 [273]3 years ago

6 0

Answer:

Power, P = 38459.16 W

Explanation:

Given that,

Mass of a car, m = 1240 kg

Initial speed, u = 0

Final speed, v = 19.4 m/s

Time, t = 10.2 s

The resistance force, R = 376 N

We need to find the average power developed by the car's engine. It is given by the formula as follows :

$P=F\times v\\\\P=(ma-R)v\\\\P=(\dfrac{mv}{t}-R)v\\\\P=(\dfrac{1240\times 19.4}{10.2}-376)\times 19.4\\\\P=38459.16\ W$

So, the required power is 38459.16 watts.

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

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Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

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Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

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Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

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The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

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Answer:

their final velocity is 0.091 m/s

Explanation:

Given;

mass of the first train, m₁ = 138,000 kg

mass of the second train, m₂ = 123,000 kg

initial velocity of the first train, u₁ = 0.288 m/s

initial velocity of the seocnd train, u₂ = -0.131 m/s (opposite direction to the first)

Let their common final velocity after been coupled = v

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(138,000 x 0.288) + (-0.131 x 123,000) = v(138,000 + 123,000)

39,744 - 16,113 = v(261,000)

23,631 = v(261,000)

v = 23,631 / 261,000

v = 0.091 m/s

Therefore, their final velocity is 0.091 m/s

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What is the primary difference between an ideal emf device and a real emf device?.

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A real emf device has an internal resistance, but an ideal emf device does not.

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Answer:

concentric with the wire

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3 years ago

A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/

german

Answer:

<h2>128.61 Watts</h2>

Explanation:

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Power = Workdone by torque/time

Workdone by the torque = $\tau \theta$ = $I\alpha * \theta$

I is the rotational inertia = 16kgm²

$\theta = angular\ displacement$

$\theta = 2 rev = 12.56 rad$

$\alpha \ is \ the\ angular\ acceleration$

To get the angular acceleration, we will use the formula;

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Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

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