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3 years ago

If Mars were 10 times closer to the Sun, then the Sun would attract Mars with

2 answers:

7 0

The velocity does not change greatly between Mars and Venus —explained Dong— but Venus's closer proximity to the sun boosts the density by almost a factor of 4.5. This would mean that atmosphere on Mars would be lost even more rapidly than at its current position.

Goshia [24]3 years ago

4 0

Answer:

Explanation:

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"a needle can be made to "float" on the surface tension of water. what causes this surface tension to form"

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The cohesion of water molecules to each other

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3 years ago

Why can the sun transmit heat to Earth only by radiation?. . The space between the sun and Earth contains almost no matter.. . R

barxatty [35]

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3 years ago

Read 2 more answers

.A cart rolling down an incline for 5.0 seconds has an acceleration of 1.6 m/s2. If the cart has a beginning speed of 2.0 m/s, a

Ilia_Sergeevich [38]

Use the formula,

$\Delta x=v_it+\dfrac12at^2$

where $\Delta x$ is the cart's displacement (from the origin), $v_i$ is its initial speed, $a$ is its acceleration, and $t$ is time.

$\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2$

$\implies\boxed{\Delta x=30\,\mathrm m}$

Alternatively, since acceleration is constant, we have

$\dfrac{v_f+v_i}2=\dfrac{\Delta x}t$

That is, we have these two equivalent expressions for average velocity, where $v_f$ is the cart's final velocity. Solve for $\Delta x$ :

$\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}$

$\implies\boxed{\Delta x=30\,\mathrm m}$

8 0

3 years ago

A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration

Dimas [21]

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

$x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}$ (1)

Where:

$x_{o}$ - Initial x-position, in meters.

$v_{o,x}$ - Initial x-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{x}$ - x-acceleration, in meters per second.

If we know that $x_{o} = 0\,m$ , $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}$

$x(t) = 0.324\,m$

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

$y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}$ (2)

Where:

$y_{o}$ - Initial y-position, in meters.

$v_{o,y}$ - Initial y-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{y}$ - y-acceleration, in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o,y} = -3.6\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{y} = 0\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is: