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Alecsey [184]
3 years ago
8

Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.

Physics
1 answer:
nydimaria [60]3 years ago
8 0

Answer:

Explanation:

Using E= hc/wavelength

6.63 x10^-34 x3x10^8/ 236nm

19.86*10^-26/236*10^-9

=0.08*10^-35Joules

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If we add 50 Joules of thermal energy to a heat engine, and that heat engine does 30 Joules of work, how much thermal energy is
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The correct answer should be

A. 20 Joules

Explanation:

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7 0
2 years ago
To find the number of neutrons in an atom, you would subtract
Luda [366]
Atomic number is equal to the number of protons and electrons

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protons + neutrons = atomic mass

I hope this helps
3 0
3 years ago
If the refractive index of a medium is 1.3,
Nataly [62]

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3 years ago
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Elemental analysis of the unknown gas from part a revealed that it is 30.45% n and 69.55% o by mass. What is the molecular formu
Anna71 [15]

Answer:

the molecular formula for the gas is NO₂

Explanation:

since it contains

Nitrogen = n → 30.45%

Oxygen = o → 69.55%

and 30.45%+69.55% = 100% , then the gas only contains nitrogen and oxygen

Also we know that the proportion of oxygen over nitrogen  is

proportion of oxygen over nitrogen  = moles of oxygen / moles of nitrogen

since

moles = mass / molecular weight

then for a sample of 100 gr of the unknown gas

mass of oxygen = 69.55%*100 gr = 69.55 gr

mass of Nitrogen = 30.45%*100 gr = 30.45 gr

proportion of oxygen over nitrogen = (mass of oxygen/ molecular weight)/(mass of nitrogen / molecular weight of nitrogen ) =  (69.55 gr/ 16 gr/mol) /( 30.45 gr /14 gr/mol) = 1.998 mol of O/ mol of N≈ 2 mol of O/ mol of N

therefore there are 2 atoms of oxygen per atom of nitrogen

thus the molecular formula for the gas is:

NO₂

6 0
3 years ago
An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele
Citrus2011 [14]

Answer:

The magnitude of the applied torque is 6.0\times10^{-2}\ N-m

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

\tau=I\times\alpha

Here, I = mk²

\tau=mk^2\times\alpha

Put the value into the formula

\tau=3\times(0.2)^2\times0.5

\tau=0.06\ N-m

\tau=6.0\times10^{-2}\ N-m

Hence, The magnitude of the applied torque is 6.0\times10^{-2}\ N-m

7 0
3 years ago
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