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pshichka [43]
3 years ago
7

What is your wheel and axle

Physics
2 answers:
Crank3 years ago
8 0

Answer:

Pls mark me brainliest

Explanation:

The wheel and axle is a simple machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.

Fittoniya [83]3 years ago
7 0

Explanation:

<em>The</em><em> </em><em>wheel</em><em> </em><em>and</em><em> </em><em>axle</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>type</em><em> </em><em>of</em><em> </em><em>simple</em><em> </em><em>machine</em><em> </em><em>used</em><em> </em><em>to</em><em> </em><em>make</em><em> </em><em>tasks</em><em> </em><em>easier</em><em> </em><em>in</em><em> </em><em>terms</em><em> </em><em>of</em><em> </em><em>manipulating</em><em> </em><em>force</em><em> </em><em>by</em><em> </em><em>applying</em><em> </em><em>the</em><em> </em><em>concept</em><em> </em><em>of</em><em> </em><em>mechanical</em><em> </em><em>advantage</em><em>. </em><em> </em>

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You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

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The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

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