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3 years ago

While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself?

1 answer:

5 0

You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).

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How is the energy of a photon related to its frequency?

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The photon's energy is equal to

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3 years ago

How does gravity correspond to the bending of the space-time fabric

Klio2033 [76]

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4 years ago

Dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks ea

Yuliya22 [10]

Answer:

The displacement is 6.71 [m} and the angle is 63.4° to the north of east

Explanation:

Using a sketch showing Dante's displacement, we can find each of the points Dante moves through. First, it moves 4 blocks east, the new coordinate (4.0), then moves 3 blocks south and we will get the new coordinate (4, -3). Then Dante moves 1 block west, thus the new point is (3, -3). And finally it moves 9 blocks north where the new coordinate in and gets -3 - (-9) = + 6.

The displacement can be found using the equation for the straight line.

$d= \sqrt{(x_{1}-x_{0} )^{2} +(y_{1}-y_{0} )^{2} } \\d= \sqrt{(3-0 )^{2} +(6-0 )^{2} } \\\\d=6.71 [m]\\$

We can realize that the triangle formed is a right triangle, therefore we can find the angle of the displacement.

$tan(a)=\frac{6}{3} \\a=tan^-1(2)\\a=63.4[deg]$

8 0

3 years ago

What is longitudinal wave?

nignag [31]

Answer: The definition is a longitudinal wave is a wave consisting of a periodic disturbance or vibration that takes place in the same direction as the advance of the wave

Explanation:

7 0

3 years ago

A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, c

Aleks04 [339]

Answer:

The velocities of the skaters are $v_{1} = 3.280\,\frac{m}{s}$ and $v_{2} = 0.024\,\frac{m}{s}$ , respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$ (1)

Second skater

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$ (2)

Where:

$m_{1}$ - Mass of the first skater, in kilograms.

$m_{2}$ - Mass of the second skater, in kilograms.

$v_{1,o}$ - Initial velocity of the first skater, in meters per second.

$v_{1}$ - Final velocity of the first skater, in meters per second.

$v_{b}$ - Launch velocity of the meter, in meters per second.

$v_{2}$ - Final velocity of the second skater, in meters per second.

If we know that $m_{1} = 70\,kg$ , $m_{b} = 0.043\,kg$ , $v_{b} = 32\,\frac{m}{s}$ , $m_{2} = 58.5\,kg$ and $v_{1,o} = 3.30\,\frac{m}{s}$ , then the velocities of the two people after the snowball is exchanged is: