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noname [10]
4 years ago
7

Jerome is learning how the model of the atom has changed over time as new evidence was gathered. He has images of four models of

the atom, but they are not in the correct order.
An image at left labeled W with overlapping red and blue balls in the center with a circular fuzzy green cloud outside them. An image at center left labeled X with a purple ball in the center surrounded by overlapping concentric black ovals, each with a small green ball on each of the 6 ovals. An image at center right labeled Y with a large black cross in a purple circle with a black line around the purple, with 10 small green balls dispersed within the purple circle. An image at right labeled Z with a purple center outlined in black with two concentric black circles around the center, the inner circle having 2 small green balls on it and the outer circle having 8 small green balls on it.

In what order should Jerome put these models to show the development from the earliest model of the atom to the most recent one?

Y, X, Z, W
Y, Z, W, X
W, Z, X, Y
W, Y, Z, X
Physics
2 answers:
adoni [48]4 years ago
6 0

Answer:

the answer is A. Y, X, Z, W

Explanation:

Stolb23 [73]4 years ago
5 0

Answer:

Y, X, Z, W

Explanation:

You know W is the most recent because it features the nucleus in the middle and the electron cloud which was shown in models after the others.

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A man climbs a wall that has a height of 8.4 meters and gave the potential energy of 4620 joules. His mass is about_____ kilogra
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Gravitational potential energy = mass x acceleration due to gravity x height
GPE=mgh
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PART ONE
Lina20 [59]

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass.  Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

4 0
3 years ago
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl
ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
  • Both movements are independent each other, due to they are perpendicular.
  • In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
  • Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

       v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)

  • Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s  (2)

  • In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
  • Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       \Delta y = v_{oy}  * t - \frac{1}{2} *g*t^{2} (3)

  • In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)

  • Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       \Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)

  • Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
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D. Riding a bike uphill
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