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3 years ago

A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .

1 answer:

6 0

How much work in J does the string do on the boy if the boy stands still?

answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero 
--------------------------------------... 
How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? 

answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. 

If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: 
4.5N + (boys negative acceleration * mass) = total force1 
work = total force1 x 11 meters 
--------------------------------------... 

How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? 

answer: same as above only reversed: 
4.5N - (boys negative acceleration * mass) = total force2 
work = total force2 x 11 meters

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lozanna [386]

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3 years ago

A football is thrown from the edge of a cliff from a height of 22 m at a velocity of 18 m/s [39degrees above the horizontal]. A

mihalych1998 [28]

Answer:

The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.

Explanation:

Given;

vertical height of the cliff, h = 22 m

velocity of the ball, u = 18 m/s at an angle 39⁰

vertical component of the velocity, $u_y = u \ sin \theta$

The time for the ball to get to the bottom of the cliff is calculated as;

h = ut + ¹/₂gt²

h = (u sinθ)t + ¹/₂ x 9.8 x t²

22 = (18 sin 39)t + 4.9t²

22 = 11.328t + 4.9t²

4.9t² + 11.328t - 22 = 0

Solve the above equation with formula method;

a = 4.9, b = 11.328, c = -22

$t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-11.328\ \ +/- \ \ \sqrt{(-11.328)^2 - 4(4.9\times -22)} }{2(4.9)}\\\\t = 1.26 \ s$

The time for the player to get to the bottom of the cliff;

Given maximum speed, Vx = 6.0 m/s and horizontal distance, X = 12 m;

$t = \frac{X}{V_x} \\\\t = \frac{12}{6} \\\\t = 2 \ s$

The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.

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2 years ago