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Advocard [28]
3 years ago
10

What are 2 pieces of evidence that support the big bang?

Physics
1 answer:
Alexxandr [17]3 years ago
7 0
Galaxies are red shifted and in 1964 to scientists discovered leftover Radiation from the Big Bang
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PLEASE HELP ASAP ON TIMED QUIZ
Ganezh [65]

Answer:

It is C and 100 percent sure about it

4 0
3 years ago
A line in the paschen series of hydrogen has a wavelength of 1880 nm. From what state did the electron originate?
NARA [144]

Answer:

n=4

Explanation:

Paschen series of hydrogen, is the series of transitions resulting from the hydrogen atom when electron jumps from a state of n\geq 4 to n_f=3.

Emission lines for hydrogen are given by:

\frac{1}{\lambda}=R_H(\frac{1}{n_f^2}-\frac{1}{n^2})

\lambda is the wavelength of the line emitted,R_H is the Rydberg constant for hydrogen, n_f is the final energy state of the electron and n is the energy state where the electron transition originated.

We have \lambda=1880nm=1880*10^{-9}m and n_f=3. Solving for n:

\frac{1}{\lambda R_H}=\frac{1}{n_f^2}-\frac{1}{n^2}\\\frac{1}{n^2}=\frac{1}{n_f^2}-\frac{1}{\lambda R_H}\\n^2=\frac{1}{\frac{1}{n_f^2}-\frac{1}{\lambda R_H}}\\n^2=\frac{1}{\frac{1}{3^2}-\frac{1}{1880*10^{-9}m(1.09737*10^7m^{-1})}}\\n^2=15.96\\n=4

6 0
4 years ago
Could someone help me with some info for an essay?I WILL REPOST FOR 100 POINTS IF YOU CAN.
polet [3.4K]

Answer:

Self-motivation is the surest way to stay focused. Nearly all the most significant tasks in life are tests for our motivation. Facing challenges and achieving success requires focus; this means we need to be responsible for our motivation. People who are motivated towards achieving their goals are focused on success. The following ideas will help you stay focused and motivated in your work and studies.

Explanation:

3 0
3 years ago
A solid conducting sphere has net positive charge and radius r = 0.800 m . At a point 1.20 m from the center of the sphere, the
Serga [27]

here we have to calculate the net positive charge present on he surface  of the conducting  sphere.

as the sphere is a conducting one the,the charge will be stored on its surface.

though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.

the electric field at outside of the sphere is E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

as E= \frac{-dv}{dr}

so V=\frac{1}{4\pi\epsilon} \frac{q}{r}

here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .

hence we have 27=\frac{1}{4\pi\epsilon} \frac{q}{1.20}

                            q=27×\frac{1}{9*10^{9} }* 1.20 coulomb

                                =3.6×10^{-9} coulomb [ans]

5 0
3 years ago
A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular
NeTakaya
Answer:
387 volts

Explanation:
Ohm's law is used to relate voltage, current and resistance.
The formula is as follows:V = I * R
where:
V is the applied voltage (measured in volts)
I is the current flowing (measured in amperes)
R is the resistance (measured in ohm)

In the given, we have:
current (I) = 9 amperes
resistance (R) = 43 ohm

Substitute with the givens in the above formula to get the voltage as follows:
V = 9 * 43
V = 387 volts

Hope this helps :)
4 0
3 years ago
Read 2 more answers
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