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tiny-mole [99]
3 years ago
14

A 66 kg person is parachuting and experiencing a downward acceleration of 2.1 m/s2. The mass of the parachute is 4.0 kg. (a) Wha

t is the upward force on the open parachute from the air?
Physics
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

539N

Explanation:

Parameters given:

Mass of person = 66kg

Mass of parachute = 4kg

Downward acceleration = 2.1 m/s²

The total force acting on the system is the sum of the force of gravity and the force due to air acting opposite gravity:

F = mg - F(a)

Where g = acceleration due to gravity

Force is given as:

F = ma

=> ma = mg - F(a)

=> F(a) = mg - ma

F(a) = m(g - a)

m is the mass of the entire sustem

m = 66 + 4 = 70kg

=> F(a) = 70(9.8 - 2.1)

F(a) = 70 * 7.7

F(a) = 539N

The upward force on the open parachute from the air is 539N

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Un Esquiador Inicia un Salto Horizontal con una velocidad Inicial de 25 m/s. La Altura al Final de la Rampa es de 80 m del punto
Evgesh-ka [11]

Responder:

Explicación:

Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;

T = 2Usin theta / 2

theta = 90 grados

U = 25 m / s

T = 25sin90 / 2 (9,8)

T = 25 / 19,62

T = 1,27 segundos

Por lo tanto, los cielos usarán 1.27 segundos en el aire.

La distancia horizontal es el rango;

Rango R = U√2H / g

R = 25√2 (80) /9,8

R = 25√160 / 9,8

R = 25 * √16,326

R = 25 * 4.04

R = 101,02 m

Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m

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They have the same speed 
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Is 250 g higher than 100 g in science
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Its the same thing. Is 250 grams more then 100 grams
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As a battery is used to charge a capacitor, does the overall charge inside the battery get smaller, greater, or stay the same?
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A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
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