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3 years ago

12

Joe is trying to soup up his dragster. He knows that the time needed for the car to go from 0 to 100 miles per hour varies inver

sely with the car's horsepower. At 200 horsepower, the car can go from 0 to 100 mph in 12 seconds. How long should the car take if he can increase the horsepower to 240 hp?.

1 answer:

Tatiana [17]3 years ago

6 0

Translating the first sentence into equation we get, t = k(1/h) where t is time in seconds, k is the constant and h is the horsepower. Substituting the values in the equation we have, 12s = k(1/200) we have a k = 2400 seconds – hp. To get the time at 240 hp we use the equation above and the constant, we get, t = (2400 seconds-hp)(1/240hp) t = 10seconds.

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Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

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3 years ago

A factory line moves parts for an automobile at a constant speed of 0.22 m/s. How long does it take the parts to move 8.5 m alon

OleMash [197]

Answer: C. 39 s

Explanation:

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Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

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Using formula of energy

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Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

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Using formula of energy

$E = \dfrac{hc}{\lambda}$

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Using formula of kinetic energy

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$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

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W = 125.44 J.

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Answer:

true

Explanation:

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