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3 years ago

Determine how much farther a person can jump on the moon as compared to the earth if take off speed and angle are the same

1 answer:

8 0

From Newton's Three Laws of Motion, derived formulas are already conveniently presented for a rectilinear motion at constant acceleration. One of its equations is

y = (Vf² - Vi²)/2a, where
y is the vertical height travelled by the object
Vf is the final velocity
Vi is the initial velocity
a is the acceleration

Now, when a man jumps, the only force acting on him is gravity pulling him down. When he reaches his maximum height, eventually his velocity will reach zero. So, Vf = 0. Suppose all parameters with subscript 1 refers to man jumping on Earth and those with subscript 2 refers to the man jumping on moon. Since initial velocity and angle is said to be the same, when we find the ratio of x₂/x₁, the terms (Vf²-Vi²) cancels out leaving us with

x₂/x₁ = a₂/a₁

It is common knowledge that gravity on Earth is 9.81 m/s². According to literature, the gravity on the moon is 1.62 m/s². Thus,

x₂/x₁ = a₁/a₂ = 9.81/1.62 = 6
x₂ = 6x₁

Therefore, the man jumping on the moon can reach 6 times higher than in Earth.

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Genrish500 [490]

Answer:

Explanation:

imagine urself on an elevator dont you feel lighter

3 0

3 years ago

When air resistance is ignored, _____ of the projectile affect(s) the range and maximum height of the projectile.

FromTheMoon [43]

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

Projectile is a missile designed to be fired from a rocket or gun.

A projectile is the object that is propelled by the application of an external force and then moves freely under the influence of gravity and air resistance.

The range is defined as the distance between the launch point and the point where the projectile hits the ground.

The height from the ground at the top most position of projectile is referred to as maximum height.

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

Learn more about maximum height click here brainly.com/question/6261898

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8 0

2 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0

3 years ago

Una caja de 5.0kg de masa se acelera desde el reposo a través del piso mediante una fuerza a una tasa de 2.0 /s2 durante 7.0s en

Nady [450]

Responder:

<h2>490 julios </h2>

Explicación:

Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;

Workdone = Fuerza * Distancia

Como Fuerza = masa * aceleración,

Workdone = masa * aceleración * distancia

Masa dada = 5.0kg, aceleración = 2.0m / s² d =?

Para obtener d, usaremos una de las leyes del movimiento,

d = ut + 1 / 2at²

u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s

d = 0 + 1/2 (2) (7) ²

d = 49m

Workdone = 5 * 2 * 49

Workdone = 490 Julios

4 0

3 years ago

A ball is dropped off the roof of a tall building. If the ball reaches the ground in 8 seconds, how tall is the building, in met

Ghella [55]

Let b be the height of the building, and y the height of the ball at time t, given by

y = b - 1/2 gt²

where g = 9.8 m/s² is the magnitude of the acceleration due to gravity.

It takes the ball 8 s to reach the ground, at which point y = 0, so that

0 = b - 1/2 (9.8 m/s²) (8 s)²

b = 1/2 (9.8 m/s²) (8 s)²

b = 313.6 m

5 0

4 years ago