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3 years ago

What 3 weather insturements used to measure weather?

Physics

2 answers:

JulijaS [17]3 years ago

8 0

Thermometer for measuring air and sea surface temperature

Barometer for measuring atmospheric pressure

Hygrometer for measuring humidity

Anemometer for measuring wind speed

Pyranometer for measuring solar radiation

brilliants [131]3 years ago

4 0

-- thermometer (temperature)

-- barometer (air pressure)

-- anemometer (wind speed)

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1. Describe what must happen to an atom to make it

liq [111]

Answer:

Explanation:

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3 years ago

What part of an atom bonds with other atoms? ? A the innermost electrons B valence electronsany C the outermost protons D any el

kaheart [24]

Ionic bonds occur when electrons are donated from one atom to another. Each type of atom forms a characteristic number of covalentbonds with other atoms. An example of that is a hydrogen atom with one electron in its outer shell forms only one bond, its out most orbital becomes filled with two electrons.
the outermost protons

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3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

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#LearnwithBrainly

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3 years ago

In which situation can you be at rest and moving at then same time

anyanavicka [17]

If you are stationary, but in/on a moving vehicle/object you can be at rest and moving at then same time.

<u>Explanation</u>:

A particle, when viewed from a given frame of reference, cannot be both at rest and in motion. However, in one frame of reference, a particle can be in motion whereas in another frame of reference the particle is in motion.

For example, if you are seated in a plane, the plane is stationary in that reference frame and the Earth moves under it, but in the reference frame of the Earth, the plane is moving concerning the Earth. When you are standing still on Earth, in your frame of reference, the Earth is stationary, and the Sun and stars move around the Earth.

However, in the frame of reference of the center of our solar system, the Earth orbits the Sun and the Sun are perturb slightly by the rest of the planets, but the rest of the galaxy orbits our solar system. Of course, in rest from our Galaxy, our solar system orbits a giant black hole at its center.

4 0

3 years ago

Can someone pleaseEeeeeeee ASAP answer this❗️❗️❗️❗️ I really need help it is my second time posting this❗️❗️❗️❗️

belka [17]

Answer:

Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?

Explanation:

So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s

$\sqrt{t } = \sqrt{77.55}$

so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull

3 0

3 years ago