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AnnZ [28]
3 years ago
6

(3) What is the weight of a 50-kg astronaut (a) on Earth (b) On the Moon ,(g=1.7m/s2), (c) on Mars (g=3.7m/s2) (d)in outer space

traveling with constant velocity.
Physics
1 answer:
artcher [175]3 years ago
5 0

Answer:

a) On Earth

490N

b) On the Moon

85N

c) On Mars

185N

d)in outer space traveling with constant velocity.

0

Explanation:

The weight is defined as:

W = mg (1)

Where m is the mass and g is the gravity

a) On Earth g = 9.8m/s^{2}

Then, equation 1 can be used:

W = (50Kg)(9.8m/s^{2})

W = 490Kg.m/s^{2}

but 1N = Kg.m/s^{2}

W = 490N

Hence, the weight of the astronaut on Earth is 490N

b) On the Moon g = 1.7m/s^{2}

W = (50Kg)(1.7m/s^{2})

W = 85N

Hence, the weight of the astronaut on the Moon is 85N

c) On Mars g = 3.7m/s^{2}

W = (50Kg)(3.7m/s^{2})

W = 185N

Hence, the weight of the astronaut on Mars is 185N

(d) in outer space traveling with constant velocity.

Tanking into consideration that the astronaut is traveling in outer space at a constant velocity, it can be concluded that the acceleration will be zero.

Remember that the acceleration is defined as:

a = \frac{v_{f} - v_{i}}{t}

Since the acceleration is the variation of the velocity in a unit of time.

Therefore, from equation 1 is gotten.      

W = (50kg)(0)

Remember that g is the acceleration that a body experience as a consequence of the gravitational field.

 

W = 0

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1. What could be the energy of a flying aeroplane?<br><br> 2. Is work the same as power?
Doss [256]

Explanation:

The total energy of an aircraft flying in the atmosphere can be calculated using equation 1. [2]

E = ½ m v2 + mgh

A Boeing 737-300 has a maximum takeoff weight of 5.65 × 104 kg, a cruise altitude of h = 10,195 m, and cruise speed of 221 m/sec. Inserting these numbers into the above equation, we obtain 7.03 GJ for the energy at cruise conditions. [3] However, the engines mounted onto the wings of the plane are required to provide additional energy per time, power, in order to keep the aircraft flying at a constant altitude and speed

Work is the energy needed to apply a force to move an object a particular distance, where force is parallel to the displacement. Power is the rate at which that work is done.

7 0
3 years ago
Un objeto que es empujado sobre una superficie plana, tiene una masa de 80 kg y dicha fuerza tiene un valor de 300 N. Calcule la
Gennadij [26K]

Responder:

A. Ff = 300 N N = 784,8 N

Explicación:

Dado

Masa del cuerpo = 80 kg

Fuerza de movimiento Fm = 300N

Dado que el cuerpo no está acelerando, la fuerza de fricción (Ff) es igual a la fuerza de movimiento que actúa sobre el cuerpo, ya que la fuerza de fricción es una fuerza de oposición, es decir, Fm = Ff

Dado que Fm = 300N, Ff = 300N

La reacción normal que actúa en el cuerpo es igual al peso.

N = W = mg

g es la aceleración debida a la gravedad

g = 9,8 m / s

N = mg

N = 80 (9,81)

N = 784,8N

Por tanto, la fuerza normal que actúa sobre el cuerpo es 784,8 N

3 0
2 years ago
A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the b
jonny [76]

Answer

given,

vertical speed of stone,v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0  

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

   v = u + a t

   v = 12 - 9.8 x 5.2

  v = -38.96 m/s

ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

c) total distance travel by the stone

  vertical distance covered by the stone

 v² = u² + 2 g h

 0 = 12² - 2 x 9.8 x h

 h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.

8 0
3 years ago
un columpio de balancin tiene una barra de 6m de longitud y en ella se sientan 2 personas,una de 60kg y otra de 40kg, calcular e
Kitty [74]

Answer:

<em>El punto de apoyo debe colocarse a 3.6 metros de la persona de 40 Kg</em>

<em>La ventaja mecánica es 1.5</em>

Explanation:

<u>Máquinas Simples</u>

Un balancín es un ejemplo de máquina simple, donde se aplica una fuerza física y ésta puede amplificarse o reducirse a voluntad cambiando la configuración física de la máquina.

En nuestro caso, el punto de apoyo o fulcro se coloca entre las dos fuerzas constituyendo una máquina de primer grado.

La situación planteada se muesta en la figura anexa. Debemos averiguar el valor de x para que las dos personas sentadas en el balancín puedan estar en equilibrio.

Para determinar el valor de x, se establece la condición de equilibrio de torques mecánicos. Ya que el balancín se asume en reposo, los torques aplicados de cada lado del mismo deben ser iguales, haciendo que el torque neto sea cero.

El torque es el producto de la fuerza por la distancia:

T = F.d

De cada extremo del balancín, se aplica una fuerza igual al peso de cada persona, es decir, llamando F1 al peso de la persona de 40 Kg y F2 al peso de la persona de 60 Kg:

F_1 = 40 kg * 9.8 m/s^2=392N\\\\F_2 = 60 kg * 9.8 m/s^2=588 N

El torque neto del balancín debe ser cero, es decir (refiérase a la figura):

F_1*x=F_2*(6-x)

Reemplazando los valores obtenidos:

392*x=588*(6-x)

Operando:

392*x+588*x=588*6

Simplificando

980*x=3528

Resolviendo

\displaystyle x=\frac{3528}{980}=3.6\ m

El punto de apoyo debe colocarse a 3.6 metros de la persona de 40 Kg, es decir, a (6 - 3.6) = 2.4 metros de la persona de 60 Kg

La ventaja mecánica se calcula como el cociente de ambas distancias

\displaystyle VM=\frac{3.6}{2.4}=1.5

La ventaja mecánica es 1.5, es decir, se amplifica la fuerza vez y media

3 0
3 years ago
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