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3 years ago

HELP PLZ

Physics

2 answers:

Usimov [2.4K]3 years ago

7 0

The push will accelerate the box at a rate of 11.78m/s²

HOW TO CALCULATE ACCELERATION:

The acceleration of the box in this question, when pushed by a force of 777N, can be calculated as follows:

Force (N) = mass (kg) × acceleration (m/s²)

Acceleration = Force ÷ mass

Acceleration = 777 ÷ 66

Acceleration = 11.78m/s²

Therefore, the push will accelerate the box at a rate of 11.78m/s²

Learn more: brainly.com/question/24372530?referrer=searchResults

Anon25 [30]3 years ago

5 0

Answer:

A = 11.77 m/s²

Explanation:

Mass of box (m) = 66kg

Force (f) = 777n

Acceleration (a) = ??

F = ma

777 = 66 · a

A = $\frac{777}{66}$ = 11.77m/s²

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Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

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Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

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