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3 years ago

6

A plane can fly at 385mph in still air. If it can fly 220 miles downwind in the same amount of time it can fly upwind at 165 mil

es, what is the velocity of the wind?

1 answer:

quester [9]3 years ago

3 0

Answer:

55 mph

Explanation:

Distance = speed × time, so time = distance / speed.

If x is the speed of the wind, then:

220 / (385 + x) = 165 / (385 − x)

Cross multiply:

220 (385 − x) = 165 (385 + x)

Distribute:

84,700 − 220 x = 63,525 + 165 x

Combine like terms:

21,175 = 385 x

Solve for x:

x = 55

The speed of the wind is 55 mph.

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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

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Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

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3 years ago

Consider a Cassegrain-focus, reflecting telescope. Images recorded at Cassegrain-focus will be:

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B. Flipped compared to what is in the sky

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3 years ago

Calculate the Force of Gravity acting on an object that has a mass of 1.3 kg. The object is on Earth so use 9.8m/s2 for the acce

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To calculate the gravitational force acting on an object given the mass and the acceleration due to gravity, use the following formula.

Fg = m • g
Fg = 1.3 kg • 9.8 m/s^2
Fg = 12.74 N or about 12.7 N.

The solution is C. 12.7 N.

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3 years ago

A square current-carrying loop is placed in a uniform magnetic field B with the plane of the loop parallel to the magnetic field

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Answer:

(A) a net torque but no net force on the loop.

Explanation:

The total force on the loop is zero because the forces on the opposite sides of the loop are equal but act in opposite directions and as a result they cancel each other out. The two forces on opposite sides to the axis of rotation each give rise to a torque about the axis of rotation. This torque is directed along the axis of rotation.

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base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to laun

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So, in the toy launcher, the energy of the compressed spring, called <u>Elastic</u> <u>Potential</u> <u>Energy (PE)</u>, transforms into the movement of the plastic sphere, called <u>Kinetic</u> <u>Energy (KE)</u>. Since total energy must be constant:

$KE_{i}+PE_{i}=KE_{f}+PE_{f}$

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

$\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}$

$\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}$

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$v^{2}=\frac{50(0.1)^{2}}{0.1}$

$v=\sqrt{50(0.1)}$

$v=\sqrt{5}$

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

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3 years ago