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Elena-2011 [213]

3 years ago

9

A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of

the moon. As seen from the tower, the moon subtends an angle of 9.26x10-3 radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.

1 answer:

lesya [120]3 years ago

8 0

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = $20.6\times 10^3\ m$

$\theta$ = The angle subtended by moon = $9.6\times 10^{-3}\ rad$

Distance traveled is given by

$s=r\times\theta$

$\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}$

$\Rightarrow s=197.76\ m$

The distance traveled by the jet is 197.76 m

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This question is incomplete because part of the content is missing; here is the complete question:

Ingrid wrote the hypothesis below. If the temperature of a liquid increases, the density of the liquid decreases because the particles move farther apart.

What are the variables in her hypothesis?

A. The independent variable is the temperature, and the dependent variable is the density.

B. The independent variable is the density, and the dependent variable is the temperature.

C. The independent variable is the temperature, and the dependent variable is the distance between particles.

D. The independent variable is the distance between particles, and the dependent variable is the temperature.

The answer to this question is A. The independent variable is the temperature, and the dependent variable is the density.

Explanation:

To begin, the variables in an experiment are the factors being studied or analyzed, which are expressed in the hypothesis. According to this, the two variables in the experiment are the temperature and the density.

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3 years ago

What are possible formulas for impulse? Check all that apply.

marissa [1.9K]

Answer:

J = FΔt

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<h3>The Impulse Formula:</h3>

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Convert the yards to km and devide your answer by 40 to get the amount of time it took to reach the receiver.
Good luck ! Hope it helped.

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Can someone please help a struggling physics student?

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<h3><u>Part A:</u></h3>

<u><em>What is the maximum height the ball will reach in the air?</em></u>

Kinematics equation used:

$v_f^2=v_i^2+2ad$ , where $v_f$ is final velocity, $v_i$ is initial velocity, $a$ is acceleration, and $d$ is distance travelled. From SI units, velocity should be in $m/s$ , acceleration should be in $m/s^2$ , and distance should be in $m$

We're given that the initial velocity is 12.0 m/s in the y-direction. At the maximum height, the vertical velocity of the ball will be 0 m/s, otherwise it would not be at maximum height. This is our final velocity.

The only acceleration in the system is acceleration due to gravity, which is approximately $9.8\:\mathrm{ m/s^2}$ . However, the acceleration is acting down, whereas the ball is moving up. To express its direction, acceleration should be plugged in as $-9.8\:\mathrm{m/s^2}$ . We have three variables, and we are solving for the fourth, which is distance travelled. This will be the maximum height of the ball.

Substitute $v_i=12$ , $v_f=0$ , $a=-9.8$ to solve for $d$ :

$0^2=12^2+2(-9.8)(d),\\0=144-19.6d,\\-19.6d=-144,\\d=\frac{-144}{-19.6}=7.34693877551\approx \boxed{7.35\text{ m}}$

<u><em>What is the velocity of the ball when it hits the ground?</em></u>

This question tests a physics concept rather than a physics formula. The vertical velocity of the ball when it hits the ground is equal in magnitude but opposite in direction to the ball's initial vertical velocity. This is because the ball spends equal time travelling to its max height as it does travelling from max height to the ground (ball is accelerating from initial velocity to 0 and then from 0 to some velocity over the same distance and time). Since the ball has an initial vertical velocity of +12.0 m/s, its velocity when it hits the ground will be $\boxed{-12.0\text{ m/s}}$ . (The negative sign represents the direction. Because velocity is a vector, it is required.)

<h3><u>Part B:</u></h3>

<u>**Since my initial answer exceeds the character limit, I've attached the first question to Part B as an image. Please refer to the attached image for the answer and explanation to the first question of Part B. Apologies for the inconvenience.**</u>

<u><em>What is the direction of the velocity of the ball when it hits the ground? Express your answer in terms of the angle (in degrees ) of the ball's velocity with respect to the horizontal direction (see figure).</em></u>

This question uses a similar concept as the second question of Part A. The vertical velocity of the ball at launch is equal in magnitude but opposite in direction to the ball's final velocity. The horizontal component is equal in both magnitude and direction throughout the entire launch, since there are no horizontal forces acting on the system. Therefore, the angle below the horizontal of the ball's velocity when it hits the ground is equal to the angle of the ball to the horizontal at launch.

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$\tan \theta=\frac{12.0\text{m/s}}{2.3\text{ m/s}}$

Take the inverse tangent of both sides:

$\arctan (\tan \theta)=\arctan (\frac{12.0}{2.3})$

Simplify using $\arctan(\tan \theta)=\theta \text{ for }\theta \in (-90^{\circ}, 90^{\circ})$ :

$\theta=\arctan(\frac{12.3}{2.3}),\\\theta =79.14989537\approx \boxed{79.15^{\circ}}$

We can express our answer by saying that the direction of the velocity of the ball when it hits the ground is $\boxed{\text{approximately }79.15^{\circ} \text{ below the horizontal}}$ or $\boxed{\text{approximately }-79.15^{\circ} \text{ to the horizontal}}$ .

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