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3 years ago

Can anyone ask me how many protons, neutrons, and electrons I need and just explain where to put them.

Physics

1 answer:

Aliun [14]3 years ago

8 0

Answer:

on the first shell (ring) there will be 2 electrons

and on the 2nd shell there will be only one electron

while in the nucleus (the middle of the diagram) there will be 4 neutrons and 3 protons

Explanation:

you can see the picture attached

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Unpolarized light with an intensity of 22.4 ????ux passes through a polarizer whose transmission axis is vertically oriented. (a

irina1246 [14]

Answer:

a) I₁ = 11.2 Lux , vertical direction , b) I₂ = 1.44 Lux

Explanation:

a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed

consequently the above the processed light has half of the incident intensity and the directional of the polarized

I₁ = I₀ / 2

I₁ = 22.4 / 2

I₁ = 11.2 Lux

is polarized in the vertical direction

b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus

I₂ = I₁ cos² θ

I₂ = 11.2 cos² 69

I₂ = 1.44 Lux

8 0

3 years ago

Sunlight is a form of electromagnetic energy.<br> a. True<br> b. False

Mama L [17]

<span>The statement, Sunlight is a form of electromagnetic energy, is true. The answer is letter A. Electromagnetic energy is a form of radiant energy that released by electromagnetic radiation. One example of an electromagnetic radiation is the visible light. And visible light can be radio waves, infrared light and X - rays. The rays of the sun are a form of visible light. It has an electromagnetic radiation of UV (ultra violet) rays. That is why the radiation at day is greater than at day due to sun’s rays. </span>

8 0

3 years ago

Read 2 more answers

A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w

Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0

4 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$