Answer:
The present day model shows a nucleus composed of protons and neutrons with cloud-like spheres of different diameters surrounding the nucleus to represent the energy levels of the electrons in the atom. Rutherford's model shows electrons orbiting the nucleus along fixed, but similar- diameter circular paths.
Both models show the composition of the nucleus at the center of the atom and the much smaller electrons at some distance from the nucleus.
Explanation:
Neither of the models does a good job of representing the relative size differences of the protons, neutrons and electrons, or the distance between the nucleus and the "electron clouds."
Assuming the earth and the sun to be perfect spheres,
Volume of the sphere = 4/3 * pi * (r**3)
Volume of the earth = 4/3 * pi * ((4000*1.609 km)**3) = 1.116 E 12 km3
Volume of the Sun= 4/3 * pi * ((7 E 5 km)**3) = 1.436 E 18 km3
Density = mass /volume
Density of earth = 6 E 24 kg / 1.116 E 12 km3 = 5.376 E 12 [kg/km3]
Density of Sun= 2 E 30 kg / 1.436 E 18 km3 = 1.392 E 12 [kg/km3]
Density of earth / Density of Sun =
5.376 E 12 [kg/km3] / 1.392 E 12 [kg/km3] = 3.86
Answer:
D) 8.0 s.
Explanation:
(The following exercise is written in Spanish. Hence, explanations will be held in such language.)
La energía requerida para elevar el ascensor es el cambio en la energía potencial gravitacional, cuya magnitud es:
![\Delta E_{g} = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (20\,m)](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bg%7D%20%3D%20%28200%5C%2Ckg%29%5Ccdot%20%5Cleft%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%2820%5C%2Cm%29)
![\Delta E_{g} = 39228\,J](https://tex.z-dn.net/?f=%5CDelta%20E_%7Bg%7D%20%3D%2039228%5C%2CJ)
Ahora, la potencia media es el resultado anterior dividido por el tiempo. Se despeja el tiempo de la ecuación:
![\Delta t = \frac{\Delta E_{g}}{\dot W}](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%20%5Cfrac%7B%5CDelta%20E_%7Bg%7D%7D%7B%5Cdot%20W%7D)
![\Delta t = \frac{39228\,J}{5000\,W}](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%20%5Cfrac%7B39228%5C%2CJ%7D%7B5000%5C%2CW%7D)
![\Delta t = 7.846\,s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%207.846%5C%2Cs)
La opción más aproximada es D.
The food package will strike the ground at 11 degrees below the horizontal.
<h3>Time for the food package to hit the ground</h3>
The time for the food package to hit the ground is calculated as follows;
h = vt + ¹/₂gt²
<em>let the initial velocity be horizontal</em>
4900 = 0(t) + (0.5 x 9.8)t²
4900 = 4.9t²
t² = 4900/4.9
t² = 1,000
t = √1,000
t = 31.62 s
<h3> Final speed of the food package when it hits ground</h3>
vf(y) = vo + gt
vf(y) = 0 + (31.62 x 9.8)
vf(y) = 309.88 m/s
<h3>Angle of projection</h3>
The horizontal component of the speed will be constant, while vertical component will change
![tan(\theta ) = \frac{V_y}{V_x} \\\\\theta = tan^{-1} (\frac{V_y}{V_x})\\\\\theta = tan^{-1} (\frac{309.88}{58.1} )\\\\\theta = 79^0](https://tex.z-dn.net/?f=tan%28%5Ctheta%20%29%20%3D%20%5Cfrac%7BV_y%7D%7BV_x%7D%20%5C%5C%5C%5C%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7BV_y%7D%7BV_x%7D%29%5C%5C%5C%5C%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7B309.88%7D%7B58.1%7D%20%29%5C%5C%5C%5C%5Ctheta%20%3D%2079%5E0)
Angle below the horizontal = 90 - 79 = 11⁰
Thus, the food package will strike the ground at 11 degrees below the horizontal.
Learn more about angle of projection here: brainly.com/question/10671136