The crate is in equilibrium. Newton's second law gives
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where
• <em>n</em> = magnitude of the normal force
• <em>mg</em> = weight of the crate
• <em>p</em> = mag. of push exerted by movers
• <em>f</em> = mag. of kinetic friciton, with <em>f</em> = 0.60<em>n</em>
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It follows that
<em>p</em> = <em>f</em> = 0.60<em>mg</em> = 0.60 (43.0 kg) <em>g</em> = 252.84 N
so that the movers perform
<em>W</em> = <em>p</em> (10.4 m) ≈ 2600 J
of work on the crate. (The <em>total</em> work done on the crate, on the other hand, is zero because the net force on the crate is zero.)
Cells that are located in the Eukarya domain contain nuclei. So there you have it, your answer is Eukarya!
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Answer:
30 m/s
Explanation:
Applying,
v = u+at................ Equation 1
Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.
From the question,
Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s
Substitute these values into equation 5
v = 0+(600×0.05)
v = 30 m/s
Hence the speed at which the ball leaves the player's boot is 30 m/s
Answer:
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Explanation:
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