That is a lunar eclipse. At night, when the Earth is between the Sun and the moon, the moon would appear to be red. Just for future reference, a solar eclipse is when the Moon is between the Sun and Earth. Speaking of which, check out the solar eclipse this August!
Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x =
x = ![\frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bm_1%20%5C%201.2%20%5C%20%2B%20m%20%5C%200.2%20%7D%7BM_2%7D)
let's calculate
x =
2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m
Answer:
Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits. Bohr built upon Rutherford's model of the atom. ... So it was not possible for electrons to occupy just any energy level.
Explanation:
The answer is C. You must divide your wavelength and your frequency to get your answer.
The emerging velocity of the bullet is <u>71 m/s.</u>
The bullet of mass <em>m</em> moving with a velocity <em>u</em> has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.
If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,
![W=Fx](https://tex.z-dn.net/?f=W%3DFx)
This is equal to the change in the bullet's kinetic energy.
![W=Fx=\frac{1}{2} m(u^2-v^2)......(1)](https://tex.z-dn.net/?f=W%3DFx%3D%5Cfrac%7B1%7D%7B2%7D%20m%28u%5E2-v%5E2%29......%281%29)
If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>
Assuming the same resistive forces to act on the bullet,
![F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)](https://tex.z-dn.net/?f=F%28%5Cfrac%7Bx%7D%7B2%7D%20%29%3D%5Cfrac%7B1%7D%7B2%7D%20m%28u%5E2-v_1%5E2%29......%282%29)
Divide equation (2) by equation (1) and simplify for v<em>₁.</em>
![\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7BFx%7D%7B2%7D%20%7D%7BFx%7D%20%3D%5Cfrac%7B%28u%5E2-v_1%5E2%29%7D%7B%28u%5E2-v%5E2%29%7D%20%5C%5C%5Cfrac%7B100%5E2-v_1%5E2%7D%7B100%5E2-10%5E2%7D%20%3D%5Cfrac%7B1%7D%7B2%7D%20%5C%5Cv_1%5E2%3D5050%5C%5Cv_1%3D71.06%20m%2Fs)
Thus the speed of the bullet is 71 m/s