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3 years ago

I need help with B on #3

Physics

2 answers:

Maksim231197 [3]3 years ago

7 0

It was Ernest Rutherford

Butoxors [25]3 years ago

6 0

The awnser is Ernest Rutherford but it was also observed by Eugen Goldstein.
( most of the credit went to Rutherford tho )

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Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, what direction and value is

lina2011 [118]

Answer:

9.8m/s^2 down (option C)

Explanation:

The only acceleration acting on this motion case in the acceleration due to gravity: 9.8 m/s^2 in the downwards direction.

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2 years ago

2CO + 2NO → 2CO2 + N2 is it balannced

natita [175]

If the same atoms appear on both sides, then it's balanced.

In this reaction, there are 4 Oxygens, 2 Carbons, and 2 Nitrogens on each side. So numerically, <em>it's balanced</em>. But I don't know enough chemistry to say whether the reaction is possible.

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3 years ago

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Objective lenses are contained in a ____________ that can be turned to put a particular objective lens in place to be used.

Masteriza [31]

Answer:

Revolving nosepiece

Explanation:

The revolving nosepiece is one of the parts of a microscope, used for holding the objective lenses. They can be turned to put a particular objective lens in place to be used in order to vary magnification.

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3 years ago

How does inertia relate to coin dropping????

vekshin1

Well Inertia means something wants to stay in place, and in reality that coin wants to stay in one place, If you placed it on an index card on a cup, and SLOWLY pulled it, it wouldn't be fast enough to overcome that force, if you pulled it quickly that coin would stay in place and drop into the cup.

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3 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago