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2 years ago

Give an example of a change that the ecosystem was not able to recover from. Can you explain why? (gizmo)

Physics

1 answer:

Tju [1.3M]2 years ago

3 0

The easter island. water receded which caused the grass to dry killing plants and making it a bare and dry island

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I will fan and give 2 medals!!!!!!

Maksim231197 [3]

Corona

I. Rare gaseous envelope

II. Irregular pearly glow

III. Total solar eclipse

IV. Surrounds darkened moon

V. Caused by diffraction

Solar Prominence

I. loop shape

II. large bright gaseous

III. luminous hydrogen gas

IV. rises above chromospheres

V. anchored to surface (sun’s surface)

Solar Flare

I. Intense High-energy

II. Causes electromagnetic disruptions

III. In solar atmosphere

IV. Varies in brightness

V. Emits radiation

Sunspots

I. Surface dark spots

II. Strong magnetic fields

III. Solar magnetic storms

IV. Visible through telescope

V. Cooler than photospheres

I did this in class I got 100%

7 0

2 years ago

The town hall building is situated close to Boojho’s house. There is a clock on the top of the town hall building which rings th

KiRa [710]

Answer:

At night we experience quietness more. At this time, sounds of object will be heard more clearer

5 0

3 years ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

3 years ago

Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.

Aleonysh [2.5K]

De broglie wavelength, $\lambda = \frac{h}{mv}$ , where h is the Planck's constant, m is the mass and v is the velocity.

$h = 6.63*10^{-34}$

Mass of hydrogen atom, $m = 1.67*10^{-27}kg$

v = 440 m/s

Substituting

Wavelength $\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m$

$1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm$

So the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0

3 years ago

A scientific law

grin007 [14]

Can have any number of exceptions as long as we know about them. Hope this helps!!! Brainleist Please!!

7 0

3 years ago