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3 years ago

Explain how the weight of an object is an application of newton's second law

1 answer:

4 0

Weight is a measure of the force of gravity acting on an object. According to Newton’s laws of motion, force is directly proportional to both mass and acceleration, and the equation for force is F=m*a, where m is mass and a is acceleration. We can use this equation to solve for weight.

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As a moon follows its orbit around a planet, the maximum grav- itational force exerted on the moon by the planet exceeds the min

Mars2501 [29]

The gravitational force exerted on the moon by the planet when the moon is at maximum distance $r_{max}$ is
$F_{min}=G \frac{Mm}{r_{max}^2}$
where G is the gravitational constant, M and m are the planet and moon masses, respectively. This is the minimum force, because the planet and the moon are at maximum distance.

Similary, the gravitational force at minimum distance is
$F_{max}=G \frac{Mm}{r_{min}^2}$
And this is the maximum force, since the distance between planet and moon is minimum.

The problem says that $F_{max}$ exceeds $F_{min}$ by 11%. We can rewrite this as
$F_{max}=(1+0.11)F_{min}=1.11 F_{min}$

Substituing the formulas of Fmin and Fmax, this equation translates into
$\frac{1}{r_{min}^2}=1.11 \frac{1}{r_{max}^2}$
and so, the ratio between the maximum and the minimum distance is
$\frac{r_{max}}{r_{min}}= \sqrt{ 1.11 }=1.05$

8 0

4 years ago

The planet earth consists of land formations, large masses of water, gases in the atmosphere, and living organisms.

andrew-mc [135]

A.
The planet contains matter and energy so it is a system.

8 0

3 years ago

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

4 years ago

An astronaut weighs 8.00x10^2 newtons on the surface of Earth. What is the weight of the astronaut 6.37x10^6 meters above the su

Orlov [11]

800/2^2=2.oo10^10^2N

8 0

4 years ago

Calcula la fuerza que tiene que realizar el brazo sobre el punto medio del mango de la pala para levantar la tierra situada en l

KiRa [710]

Explanation:

1- El primer paso para calcular la fuerza del brazo en el punto medio de la excavadora para levantar la tierra, es averiguar dónde está el fulcro, que en este caso está en las manos.

2- El segundo paso es calcular el brazo de potencia, al que llamaremos Bp, y el brazo de resistencia, que estará representado por Br.

Si Bp es igual a la distancia de la potencia al fulcro, Bp = 60cm

y si Br es igual a la distancia desde la resistencia al fulcro, entonces

Br = 40 + 60 = 100cm

3- Ahora aplicamos la fórmula de la ley de la palanca. Nosotros

piden que se aplique la fuerza en manos de

en el medio, es decir, la potencia P. La fórmula sería:

Y en el tercer paso, aplicaremos la fórmula de la ley de la palanca, que es:

P = R x Br / Bp

P = 8 x 100/60

P = 13,33 Kgf

<u>Entonces la respuesta final es:</u>

P = 13,33 Kgf

7 0

3 years ago