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Lady_Fox [76]
3 years ago
8

The engine on a fighter airplane can exert a force of 105,840 N (24,000 pounds). The take-off mass of the plane is 16,875 kg. (I

t weighs 37,500 pounds.) If you mounted this aircraft engine on your car, what acceleration would you get? (Please use metric units. The data in pounds are given for comparison. Use a reasonable estimate for the mass of your car. A kilogram mass weighs 2.2 pounds.)
Physics
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

The acceleration you can get with that engine in your car is around 70,56 (\frac{m}{s^{2} }) or 7,26 (\frac{ft}{s^{2} } ) using 1500kg of mass or 3306 pounds

Explanation:

Using the equation of the force that is:

F=m*a

So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds

a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }

a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }

<em> Note: N or Newton units are: \frac{kg * m}{s^{2} }</em>

a= 70,54 \frac{m}{s^{2} }

Also in pounds you can compared

a= \frac{2400  lf }{3 306  lf}

Note: lf in force units are: \frac{lf*ft}{s^{2} }

a=7,26 \frac{ft}{s^{2} }

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Answer:

Part a)

10\hat i + 15\hat j = \vec v

Part b)

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Explanation:

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m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

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5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v

10\hat i + 15\hat j = \vec v

Part b)

magnitude of the initial speed of A = \sqrt{15^2 + 30^2} = 33.54 m/s

magnitude of the initial speed of B = \sqrt{10^2 + 5^2} = 11.18 m/s

magnitude of final speed of A = \sqrt{5^2 + 20^2} = 20.61 m/s

magnitude of final speed of B = \sqrt{10^2 + 15^2} = 18.03 m/s

Now change in total kinetic energy is given as

\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)

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A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
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Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

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E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

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