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1 year ago

FInal physics question please help asap

Physics

1 answer:

blondinia [14]1 year ago

5 0

The cannonball will be 311.8 m far in meter when it travels horizontally

<h3>What is a Projectile ?</h3>

A projectile is any object like stone, ball or javelin that take a parabolic path when thrown into the air.

The following parameters are given from the question

Ф = 30°

u = 60 m/s

g = 10 m/s²

R = ?

Resolve the initial velocity into horizontal and vertical components. That is,

Horizontal component = ucosФ

Vertical component = usinФ

To calculate the range that is, the horizontal distance, will can first calculate the total time take by the ball to reach the ground.

Total time T = 2usinФ / g

Substitute all the parameters

T = (2 x 60sin30) / 10

T = (2 x 30) / 10

T = 60 / 10

T = 6 s

The horizontal distance R = ucosФ x T

R = 60 x cos 30 x 6

R = 311.77 m

Therefore, the cannonball will travel 311.8 m approximately horizontally.

Learn more about Projectile here: brainly.com/question/24949996

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Using dimensional analysis, construct a constant, with units of length only, out of all three of the following fundamental const

ludmilkaskok [199]

The quantity with units of length only is $\sqrt{\frac{hG}{c^3}}$

Explanation:

We have to combine the following constants:

- $h$ , Planck constant, with units $[m^2][kg][s^{-1}]$

- G, the Newton's gravitational constant, with units $[m^3][kg^{-1}][s^{-2}]$

- $c$ , the speed of light, with units $[m][s^{-1}]$

The combination of these constant should have units of length only, so with meters (m).

First, we notice that $h$ has [kg] in its units, while G has $[kg^{-1}]$ in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

$hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]$

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by $c^3$ , so that the new units are:

$\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]$

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

$\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]$