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3 years ago

A soccerball is kicked straight out from a hill at 15 m/s and lands 42m away. how tall is the hill

Physics

1 answer:

ozzi3 years ago

4 0

Answer:

The height of the hill is, h = 38.42 m

Explanation:

Given,

The horizontal velocity of the soccer ball, Vx = 15 m/s

The range of the soccer ball, s = 42 m

The projectile projected from a height is given by the formula

S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

h = S²g/2Vx² (Since Vy = 0)

Substituting the values

h = 42² x 9.8/ (2 x 15²)

= 38.42 m

Hence, the height of the hill is, h = 38.42 m

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$\epsilon = NBA\omega$

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N = Number of loops

B = Magnetic Field

A = Cross-sectional Area

$\omega$ = Angular velocity

Re-arrange to find N,

$N = \frac{\epsilon}{BA\omega}$

Our values are given as,

$\epsilon = 19V$

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$N = \frac{\epsilon}{( 0.434)A\omega}$

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6 0

3 years ago

How many butter, which has a usable energy content of 6.0 Cal/g (= 6000 cal/g), would be equivalent to the change in gravitation

Mazyrski [523]

Answer:

175.96 g

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= 1.0557 x 10⁶ cals

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= 175.96 g of butter.

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2 years ago

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3 years ago

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rewona [7]

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$V = \sqrt{\frac{T}{\rho}}$

Where,

T = Tension

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Our data are given by

Tension , T = 70 N

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Amplitude , A = 7 cm = 0.07 m

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$V = \sqrt{\frac{T}{\rho}}$

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snow_tiger [21]

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