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2 years ago

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A moto-cross motorcycle/rider is airborne after a jump. Each wheel of his bike has a rotational inertia of 0.760 kg m2, so the p

air have twice that. He is moving at a speed of 15.0 m/s, so the tires are spinning at 50.0 radians/s. He and the motorcycle have a rotational inertia of 35.0 kg m2. In the air he panics, and clamps on the brake so the tires stop compared to the bike. What will his (bike and all) angular speed be after the wheels stop

1 answer:

Debora [2.8K]2 years ago

6 0

Answer:2.17 rad/s

Explanation:

Given

Each wheel has moment of inertia $I=0.76 kg-m^2$

Linear speed speed $v=15 m/s$

angular speed of tire $\omega =50 rad/s$

moment of inertia of motorcycle $I_2=35 kg-m^2$

Initial moment of inertia $I_1=2\times 0.76=1.52 kg-m^2$

Due to absence of external Torque we can Conserve Angular Moment of inertia

$l_1\omega _1=I_2\cdot \omega _2$

$2\times 0.76\times 50=35\times \omega _2$

$\omega _2=2.17\ rad/s$

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2 years ago

A ball rolls 6.0 meters as its speed changes from 15 meters per second to 10 meters per second. What is the average speed of the

antoniya [11.8K]

Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m

The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²

The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s

The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s

Answer: 12.5 m/s

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2 years ago

At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat

vlabodo [156]

Answer:

$20.3125$ kJ/mol

Explanation:

$P_{i}$ = initial vapor pressure = 45.77 mm Hg

$P_{f}$ = final vapor pressure = 193.1 mm Hg

$T_{i}$ = initial temperature = 213.1 K

$T_{f}$ = final temperature = 243.7 K

$H$ = Heat of vaporization

Using the equation

$ln\left ( \frac{P_{f}}{P_{i}} \right ) = \left ( \frac{-H}{R} \right )\left ( \frac{1}{T_{f}} - \frac{1}{T_{i}}\right)$

$ln\left ( \frac{193.1}{45.77} \right ) = \left ( \frac{-H}{8.314} \right )\left ( \frac{1}{243.7} - \frac{1}{213.1}\right)$

$H = 20312.5$ J/mol

$H = 20.3125$ kJ/mol

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2 years ago

At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti

dalvyx [7]

Answer:

Part(a): The angular acceleration is $5.63~rad~s^{-2}$ .

Part(b): The angular displacement is $2629~rad$ .

Explanation:

Part(a):

If $\omega_{1},~\omega_{2}~and~\alpha$ be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

$\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)$

where ' $t$ ' is the time taken by the centrifuge to increase its angular speed.

Given, $\omega_{i} = 250~rad~s^{-1}$ , $\omega_{f} = 750~rad~s^{-1}$ and $t = 9.5~s$ . From equation ( $I$ ), the angular acceleration is given by

$\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}$

Part(b):

Also the angular displacement ( $\Delta \theta$ ) can be written as

$&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad$

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2 years ago

Convert the volume 8.06 in.3 to m3, recalling that1in. =2.54cmand100cm=1m. Answer in units of m3.

galina1969 [7]

1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
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Therefore:

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Answer: 8.06 in³=1.321 x 10⁻⁴ m³

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2 years ago