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Gala2k [10]
3 years ago
5

An atom’s emission of light with a specific amount of energy confirms that

Physics
2 answers:
Vlad1618 [11]3 years ago
6 0

The answer is...

B-electrons emit and absorb energy based on their position around the nucleus.

I just took the test

laila [671]3 years ago
3 0
<span>An atom’s emission of light with a specific amount of energy confirms that </span><span>electrons emit and absorb energy based on their position around the nucleus.

The light emitted from an electron is a result of the electron's quantum jumps/leaps ( atomic electron transitions ) to and from different energy levels.</span>
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A horizontal force of 350N is exerted on a 2.5 kg ball as it rotates uniformly in a horizontal circle of radius of 0.90m. Calcul
harkovskaia [24]
F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s
5 0
3 years ago
True or false this is for a test
OleMash [197]

Answer:

true

Explanation:

7 0
3 years ago
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A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to
dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence,i=30^{\circ}

Angle of refraction,r=19.24^{\circ}

a.Refractive index of air,n_1=1

We know that

n_2sinr=n_1sini

n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517

b.Wavelength of red light in vacuum,\lambda=632.8nm=632.8\times 10^{-9} m

1nm=10^{-9} m

Wavelength in the solution,\lambda'=\frac{\lambda}{n_2}

\lambda'=\frac{632.8}{1.517}=417nm

c.Frequency does not change .It remains same in vacuum and solution.

Frequency,\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}

Where c=3\times 10^8 m/s

Frequency,\nu=4.74\times 10^{14}Hz

d.Speed in the solution,v=\frac{c}{n_2}

v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s

5 0
3 years ago
A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency
Mice21 [21]

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

\mu = \frac{m}{L}

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

\mu = \frac{0.09\ kg}{1\ m}

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

<u>T = 712.9 N</u>

7 0
3 years ago
Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ
elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level =  KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

5 0
3 years ago
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