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mezya [45]
1 year ago
5

a 1300-kilogram space vehicle travels at 4.8 meters per second along the level surface of mars. if the magnitude of the gravitat

ional field strength on the surface of mars is 3.9 newtons per kilogram, the magnitude of the normal force acting on the vehicle is (1) 320 n (2) 4400 n (3) 930 n (4) 5070 n
Physics
1 answer:
Helga [31]1 year ago
5 0

The magnitude of the normal force acting on the space vehicle will be 5070 newtons. So option (4) is correct.

Given that,

The mass of the space vehicle is 1300 kilograms

It travels at a speed of 4.8 meters per second along the level surface of mars.

The g' (gravitational field strength) on the surface of mars is 3.9 newtons per kilogram

We need to find the normal force acting on the space vehicle

We know, N = mg' (where N is normal force)

N = (1300 × 3.9) Newton

N = 5070 Newton

So, the magnitude of the normal force acting on the space vehicle that is traveling at 4.8 meters per second along the level surface of mars will be 5070 newtons.

The normal force is the force that surfaces exert to prevent solid objects from passing through each other. The normal force is one type of ground reaction force.

Learn more about the normal force here:

brainly.com/question/6922206

#SPJ4

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6 0
2 years ago
Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s
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Answer:

42m/s

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Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}

with this value of vo you calculate the max time:

t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

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