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3 years ago

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A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an x axis. He can legally

pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive x component. Suppose the player runs at speed 3.5 m/s relative to the field while he passes the ball with velocity
v⃗ BP
relative to himself. If
v⃗ BP
has magnitude 6.0 m/s, what is the smallest angle it can have for the pass to be legal?

1 answer:

Katyanochek1 [597]3 years ago

6 0

Answer:125.68°

Explanation:

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ou are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soo

madam [21]

Till the time car is just adjacent to the bicycle we can say

distance moved by cycle = distance moved by car

Time taken by car to accelerate from rest

$t = \frac{v_f - v_i}{a}$

$t = \frac{49 - 0}{7} = 7 s$

Time taken by cycle to accelerate

$t = \frac{23 - 0}{15} = 1.53 s$

now the distance moved by cycle in time "t"

$d = \frac{23 + 0}{2}*1.53 + 23(t - 1.53)$

distance moved by car in same time

$d = \frac{7t + 0}{2}(t)$

now make them equal

$3.5t^2 = 17.595 - 35.19 + 23t$

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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

<u>Using equation of motion:</u>

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

A)

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

B)

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

C)

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

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Margarita [4]

A. The formula for mean free time is:

t = V/(4π√2 r²vN)
where
N = 1×10¹⁶ molecules (per m³)
V = 1 m³
r = 111×10⁻⁷m (atomic radius of silicon)

Let's solve for v first:
v = √(3RT/M) = √(3(8.314 m³·Pa/mol·K)(25 + 273 K)/28.1 g/mol Si)
v = 16.26 m/s

t = (1 m³)/(4π√2 (111×10⁻⁷m)²(16.26 m/s)(1×10¹⁶ molecules))
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