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3 years ago

9

6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t

he tee. A golfer hits a ball at an angle of 8.6° above the horizontal and makes a hole-in-one! What was the initial speed of the ball?

1 answer:

Georgia [21]3 years ago

8 0

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of 12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$ ................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$ ..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

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Explanation:

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2 years ago

You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul

photoshop1234 [79]

1) First of all, let's find the resistance of the wire by using Ohm's law:
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4 years ago

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LuckyWell [14K]

Answer:

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Explanation:

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2) For the temperature of wood to raise at any point to the next degree by 1°C, will require a specific heat capacity of 1.760 J/Kg°C

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3 years ago

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lutik1710 [3]

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3 years ago

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Volgvan

Answer:

A=0.199

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4 0

3 years ago