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2 years ago

My brainly won’t load please help it’s glitching and saying “oops, something went wrong”

Physics

1 answer:

dmitriy555 [2]2 years ago

3 0

Answer:

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Explanation:

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Si un vector tiene una dirección de 2300 a partir del eje x positivo, ¿Qué signos tendrán sus componentes x y y? Si la la razón

Masteriza [31]

If you can write this in English I can help

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2 years ago

he acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. The

Ahat [919]

1/3 the weight than it is on earth

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3 years ago

The weight of an object on Earth is 350 newtons. On Mars, the same object would weigh 134 newtons. What is the acceleration due

sineoko [7]

W = mg = 350 newton
m = W/g = 350/9.8 = 35.71 kg
on mars
W = mg = 134 newton
g = W/m = 134/35.71 = 3.75 meters/second2

4 0

2 years ago

Read 2 more answers

What does Newton's 2nd Law explain?

kow [346]

Answer:

C. Why you must push harder to move a car farther.

Explanation:

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

$Acceleration = \frac {Force}{mass}$

Hence, Newton's 2nd Law explains why you must push harder to move a car farther because of its mass. Thus, it is important to increase the force that the engine provides and decrease the mass of the car.

6 0

2 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

2 years ago