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nirvana33 [79]
3 years ago
13

A place kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear

the crossbar, which is 3.05 m high. when kicked, the ball leaves the ground with a speed of 23.4 m/s at an angle of 50.0° to the horizontal. (a) by how much does the ball clear or fall short of clearing the crossbar
Physics
1 answer:
trapecia [35]3 years ago
4 0
<span>The ball clears by 11.79 meters Let's first determine the horizontal and vertical velocities of the ball. h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s Now determine how many seconds it will take for the ball to get to the goal. t = 36.0 m / 15.04 m/s = 2.394 s The height the ball will be at time T is h = vT - 1/2 A T^2 where h = height of ball v = initial vertical velocity T = time A = acceleration due to gravity So plugging into the formula the known values h = vT - 1/2 A T^2 h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2 h = 42.92 m - 4.9 m/s^2 * 5.731 s^2 h = 42.92 m - 28.0819 m h = 14.84 m Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
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latent heat of fusion =33.5 x 10^(4) J/kg

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for the water to be converted to ice it must undergo three stages:

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Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

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