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3 years ago

A Long Jumber leaves the ground at on

Physics

1 answer:

MArishka [77]3 years ago

7 0

Answer:

horizontal velocity vh = 6*cos(30°) = 6*(√3)/2 = 3√3 m/s

initial vertical velocity vv = 6*sin(30°) = 6/2 = 3m/s

Using s = ut + at2/2 for change in vertical distance in time t, with acceleration a (-9.8m/s2) and initial velocity u (vv = 3m/s) we have

0 = 3*t - 9.8*t2/2 or t = 6/9.8 s (ignoring the t = 0 solution, which just represents staying still!).

The horizontal distance in time t is vh*t or 3√3*6/9.8 m

Explanation:

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2 years ago

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9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en

masha68 [24]

Answer:

2.61 atm

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Explanation:

$P_1$ = Presión inicial = 0.96 atm

$P_2$ = Presión final

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$V_2$ = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

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3 years ago

A car is moving at 76 miles per hour. the kinetic energy of that car is 5 × 105 j. how much energy does the same car have when i

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3 years ago

A penny is dropped from the Statue of Liberty

Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

$vf=vi+at$ ,

vf =unknown velocity vi= initial velocity vi=0m/s a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

$vf=0m/s+(9.8m/s^2)*0.30seconds$

Step 2: Solve

$vf=2.94m/s$

Use the equation below to solve for the displacement at 0.30 seconds

$x=vit+\frac{1}{2} at^{2}$

Step 1: Substitute the same variables with the knowns

$x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}$

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

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3 years ago