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kicyunya [14]
3 years ago
10

The Valence electrons determine an atom's ?

Physics
1 answer:
olga55 [171]3 years ago
3 0

Answer:

The valence electrons determine an atom's reactivity.

Explanation:

The valence electrons is the electrons on the most outer layer of an atom. The higher the number of valence electrons, the bigger the reactivity of the atom. This means that when that atom is put with another substance and it reacts, it would react in a much more violent way than another atom with lower valence electrons.

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Find the frequency, if the amplitude of a 3000g object in simple harmonic motion is 1000cm and the maximum speed of the object i
Oksana_A [137]

Answer:

A = 10 m     amplitude

m = 3 kg     mass of object

Vm = 5 m/s

w A = Vm      where w = omega

w = 2 * pi * f

2 * pi * f  10 = 5

f = 5 / (20 * pi)  = .0796 / sec

7 0
2 years ago
Help 30 points and willgive brainliest
MissTica
HEJEGEJDHIEBDJDHDIDGDJGJDHD
8 0
2 years ago
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
Strike441 [17]

Answer:

E=3.5(8.98*10^{6}x-2.69*10^{15}t)

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)

Explanation:

The electric field equation of a electromagnetic wave is given by:

E=E_{max}(kx-\omega t) (1)

  • E(max) is the maximun value of E, it means the amplitude of the wave.
  • k is the wave number
  • ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

k=\frac{2\pi}{\lambda}            

k=8.98*10^{6} [rad/m]      

And the relation between λ and f is:                

c=\lambda f

f=\frac{c}{\lambda}

f=\frac{3*10^{8}}{700*10^{-9}}

f=4.28*10^{14}

The angular frequency equation is:

\omega=2\pi f

\omega=2\pi*4.28*10^{14}

\omega=2.69*10^{15} [rad/s]

Therefore, the E equation, suing (1), will be:

E=3.5(8.98*10^{6}x-2.69*10^{15}t) (2)

For the magnetic field we have the next equation:

B=B_{max}(kx-\omega t) (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

E_{max}=cB_{max}

B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}

B_{max}=1.17*10^{-8}T

Putting this in (3), finally we will have:

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t) (4)

I hope it helps you!

8 0
3 years ago
The average power dissipated in a 47 ω resistor is 2.0 w. what is the peak value i 0 of the ac current in the resistor?
pishuonlain [190]
The average dissipated power in a resistor in a ac circuit is:
P=I_{rms}^2 R
where R is the resistance, and I_{rms} is the root mean square current, defined as
I_{rms} =  \frac{I_0}{\sqrt{2}}
where I_0 is the peak value of the current. Substituting the second formula into the first one, we find
P=( \frac{I_0}{\sqrt{2} } )^2 R =  \frac{1}{2} I_0^2 R
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
I_0 =  \sqrt{ \frac{2 P}{R} }=  \sqrt{ \frac{2 \cdot 2.0 W}{47 \Omega} }=0.29 A
4 0
3 years ago
An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro
svet-max [94.6K]
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
6 0
3 years ago
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