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2 years ago

A ball is fired from a cannon at point 1 and follows the trajectory shown in the figure below. Air

Physics

1 answer:

AveGali [126]2 years ago

3 0

In trajectory, acceleration vector is not the same with the velocity vector but it depends on the change in velocity. The acceleration path is perpendicular to the velocity vector path. The Correct answers are:

a.) D

b.) B

c.) C

d.) A

A Vector is a quantity that has both magnitude and direction.

Given that a ball is fired from a cannon at point 1 and follows the trajectory shown in the figure.

a.) The vector best represents the ball's velocity at position 2 is D

b.) The vector best represents the ball's acceleration at point 2 is B

c.) Which vector best represents the ball's velocity at position 3 = C

d.) Which vector best represents the ball's acceleration at point = A

In trajectory, acceleration vector is not the same with the velocity vector but depends on the change in velocity. The acceleration path is perpendicular to the velocity vector path.

At point 3, since the ball path is parabolic, the velocity at that point is not equal to zero but will tend to take a horizontal path.

Learn more here: brainly.com/question/25519989

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ohaa [14]

Answer:

2.83 m

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is given by

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

For the FM radio waves in this problem, we have:

$f_1=89 MHz=89\cdot 10^6 Hz$ is the minimum frequency, so the maximum wavelength is

$\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m$

The maximum frequency is instead

$f_2=106 MHz=106\cdot 10^6 Hz$

Therefore, the minimum wavelength is

$\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m$

So, the wavelength at the beginning of the range is 2.83 m.

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Kruka [31]

You can’t solve it because you don’t have c in the question

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Musya8 [376]

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Explanation:

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Work done = Force x distance

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