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3 years ago

???help??? \ / lnl HELP PLS

Physics

2 answers:

Vinil7 [7]3 years ago

4 0

It comes from the word moon

xxTIMURxx [149]3 years ago

3 0

Answer:

It comes from the word "moon" and it is because it takes a full moon cycle around the earth to complete a month, around 30 days

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Two large metal plates of area 1.2 m^2 face each other, 6.5 cm apart, with equal charge magnitudes but opposite signs. The field

andrey2020 [161]

Answer:

Q= 722.5 *10⁻¹² C

Explanation:

Conceptual analysis

For a parallel plate capacitor, we can use the following formula :

E= (Q) /(ϵ₀*A) Formula (1)

Where:

E: electric field between the plates ( N/C)

Q: Charge of the plates (C)

ϵo : vacuum permittivity ( C²/ N.m²)

A : area oh the plates (m²)

Known data

A = 1.2 m²

E= 68 N/C

ϵo= 8.8542*10⁻¹² ( C²/ N.m²)

Problem development

We apply the formula (1) :

$68= \frac{Q}{(8.8542*10^{-12} )(1.2)}$

Q= (68) (8.8542*10⁻¹²)(1.2)

Q= 722.5 *10⁻¹² C

6 0

3 years ago

How can doing work on an object increase its kinetic energy?

Lyrx [107]

Answer:If the object's speed increases.

Explanation:

If the object's speed increases, then its kinetic energy will increase. If the kinetic energy increases, the change in kinetic energy will be positive.

6 0

3 years ago

a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s

Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

$h = v_0_yt + \frac{1}{2} gt^2$

where;

t is the time of motion
$v_0_y$ is the initial vertical velocity of the stone = 0

$h = \frac{1}{2} gt^2$

The time taken by the first stone to hit the ground is calculated as;

$t_1 = \sqrt{\frac{2h}{g} }$

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

$t_2 = \sqrt{\frac{2h}{g} } + 1$

$t_2 = t_1 + 1$

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"Your question is not complete, it seems be missing the following information;"

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0

3 years ago

You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb

mafiozo [28]

Answer:

The angle it subtend on the retina is $\theta_z = 0.44586^o$

Explanation:

From the question we are told that

The length of the warbler is $L = 14cm = \frac{14}{100} = 0.14m$

The distance from the binoculars is $d = 18cm = \frac{18}{100} = 0.18m$

The magnification of the binoculars is $M =8$

Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as

$\theta = \frac{L}{d}$

$\theta = \frac{0.14}{0.18}$

$= 0.007778 rad$

Now magnification can be represented mathematically as

$M = \frac{\theta _z}{\theta}$

Where $\theta_z$ is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.

$\theta_z = M * \theta$

=> $\theta_z =8 * 0.007778$

$= 0.0622222224$

Generally the conversion to degrees can be mathematically evaluated as

$\theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )$

$\theta_z = 0.44586^o$

7 0

3 years ago

A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon

Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

$n=\frac{m}{M_m}$

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

$n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol$

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

$N=n N_A$

where

n is the number of moles

$N_A=6.022\cdot 10^{23} mol^{-1}$ is the Avogadro number

Substituting,

$N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18}$ atoms

The energy emitted by each atom (the energy of one photon) is

$E_1 = \frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda=508 nm=5.08\cdot 10^{-7}nm$ is the wavelength

Substituting,

$E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J$

And so, the total energy emitted by the sample is

$E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J$

4 0

3 years ago