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3 years ago

PLEAS HELP.

1 answer:

8 0

A: put an atom on a poster in the exhibit
 Good luck. The poster itself is made of trillions of trillions of trillions
 of atoms. You could not see the extra one any easier than you could
 see the ones that are already there, and even if you could, it would be
 lost in the crowd.

B: use a life size drawing of an atom
 Good luck. Nobody has ever seen an atom. Atoms are too small
 to see. That's a big part of the reason that nobody knew they exist
 until less than 200 years ago.

D: set up a microscope so that visitors can view atoms
 Good luck. Atoms are way too small to see with a microscope.

C: Display a large three dimensional model of an atom.
 Finally ! A suggestion that makes sense.
 If something is too big or too small to see, show a model of it
 that's just the right size to see.

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Why are some areas in Europe seeing a partial solar eclipse while others are observing a total solar eclipse?

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Answer:

Objects between the source and observer produce possible shadows known as the umbra and the penumbra.

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The penumbra is a partially shadowed are where some of the source light travels directly to the observer an some of the source light is blocked from the observer (the object blocking the light is not a point object).

The question apparently refers to the differences seen by the observers.

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3 years ago

Work is done on an object only if the force and displacement are __________?

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3 years ago

Which of the following best describes the difference between speed and acceleration?

Nataly_w [17]

The correct answer to the question is : C) Speed is the distance an object travels within a specific unit of time, whereas acceleration is the rate at which the speed or direction of an object is changing.

EXPLANATION :

Before going to answer this question, first we have to understand speed and acceleration.

The speed of a body is defined as the rate of change of distance or the distance covered by the body per unit time. Speed is scalar quantity. Hence, it needs only magnitude for its complete specification.

The acceleration produced by a body is defined as the rate of change of velocity. We know that velocity is the speed in a particular direction. Hence, the acceleration is not only produced due to the change in magnitude of velocity as well as due to the change in direction. Hence, acceleration is a vector quantity which needs magnitude as well as direction for its complete specification.

Hence, the best difference between speed and acceleration is the third statement.

3 0

3 years ago

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As a person pushes a box across a floor, the energy from the person’s moving arm is transferred to the box, and the box and the

san4es73 [151]

Answer:

conserved

Explanation:

During this process the energy is conserved

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3 years ago

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In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago