How many protons are in hydrogen -2

1 answer:

5 0

The number of protons in the nucleus of an atom determines
what element the atom is an atom of.
If there's one proton in the nucleus of an atom, then the atom
is an atom of Hydrogen.
If there's any other number of protons other than one in the nucleus
of an atom, then the atom is an atom of some other element. It's not
an atom of Hydrogen.

If you say you have an atom of Hydrogen, then I know there's exactly
one and only one proton in its nucleus. If you call it Hydrogen-2, I know
you must be talking about the Atomic Weight of the nucleus, and there
must be a neutron in there too.

If you called it Hydrogen-37, I would know that there would still be exactly
one proton there, lost in a crowd of 36 neutrons.

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On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

Anettt [7]

Answer:

The pressure difference is 25.8kPa while the force exerted on the window by air pressure is 3.02 kN

Explanation:

Using the Bernoulli's equation

$P_{in}+\frac{1}{2}\rho v_{in}^2=P_{out}+\frac{1}{2}\rho v_{out}^2$

Here

Pin-Pout is the pressure difference which is to be calculated.

ρ is the density of air whose value is 1.29 kg/m^3

vin is the velocity of air inside which is 0 m/s

vout is the velocity of air outside which is 200 m/s

Substituting values in the above equation yields

$\Delta P=P_{in}-P_{out}\\\Delta P=\frac{1}{2}\rho (v_{in}^2-v_{out}^2)\\\Delta P=\frac{1}{2}\times 1.29 ( 200^2-0)\\\Delta P=25.8 kPa$

So the pressure difference is 25.8kPa.

As force is given by

$F=PA\\F=\Delta P A\\$

Here

ΔP is the pressure difference calculated above
A is the area of the window given as

$A=L \times W\\A=0.26 \times 0.45\\A=0.117 m^2$

Now force is

$F=\Delta P A\\F=25.8 \tiems 10^3 \times 0.117\\F=3.02 kN$

So the force exerted on the window by air pressure is 3.02 kN

6 0

2 years ago

How did she change the circuit?

klio [65]

Answer:

She changed a closed circuit to an open circuit

Explanation:

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2 years ago

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If the bat emits a sound at a frequency of 80.2 kHzkHz and hears it reflected at a frequency of 83.5 kHzkHz while traveling at a

solong [7]

To solve this problem we will apply the concepts related to the Doppler Effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other.

Mathematically it can be described as

$v' = v (\frac{f' ( v - v_{bat}) - f ( v + v_{bat})}{f'(v-v_{bat}) + f ( v + v_{bat})})$