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Angelina_Jolie [31]
3 years ago
6

If gravity between the Sun and the Earth suddenly vanished, Earth would continue moving

Physics
1 answer:
zmey [24]3 years ago
4 0

Answer:

D, if the sun was removed, all the planets would go flying off into space.

Explanation:

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An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value
Dimas [21]

Answer:

n_f=2

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

R = Rydberg constant, R=1.097\times 10^7\ m^{-1}

\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

Solving above equation we get the value of final n is,

n_f=2.04

or

n_f=2

So, it will relax in the n = 2. Hence, this is the required solution.        

6 0
3 years ago
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A scientist is studying a sample of matter the matter has no definite shape or volume and it can conduct an electric current. Th
Anna35 [415]
D. plasma ..................
4 0
3 years ago
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In many cartoon shows, a character runs off a cliff, realizes his predicament, and lets out a scream. He continues to scream as
Naddik [55]

Answer:

d. lower than the original pitch and decreasing as he falls.

Explanation:

As per the Doppler effect when the pitch of the sound increases as the source approaches the observer and decreases as the source moves away. A classic example of this increasing pitch of ambulance siren as it approaches you and decreasing pitch of the siren as it goes away from you.

The same effect is applicable here as well. As the character keep falling, it is moving away from the observer so the pitch of his scream will keep on decreasing.

8 0
3 years ago
What is the purpose of the background research step of the scientific method?
Lilit [14]
Background research is used to familiarize yourself with the topic or problem you're conducting your experiment on and help you form an educated guess explaining or solving it that will become your hypothesis. 
3 0
2 years ago
Two small nonconducting spheres have a total charge of 90.0 C.
valentina_108 [34]

Answer: (a) Smaller charge is 2.7 \times 10^{-5} C and larger charge is 11.7 \times 10^{-5} C.

(b) Smaller charge is -11.4 \times 10^{-5} and larger charge is 9.1 \times 10^{-5}.

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

        Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}

                      = 9 \times 10^{-5} C

Therefore, force between the two spheres will be calculated as follows.

        F = k\frac{Q_{1}Q_{2}}{r^{2}}

       12 N = \frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}

       Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}

or,     Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}

    9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}

    Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0

        Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C

This means that smaller charge is 2.7 \times 10^{-5} C and larger charge is 11.7 \times 10^{-5} C.

(b)  When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

      Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}

      Q_{1} - Q_{2} = 9 \times 10^{-5} C

Now, force between the two spheres is calculated as follows.

    F = k\frac{Q_{1}Q_{2}}{r^{2}}

    12 N = \frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}

   Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}

   Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}

    Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}

        Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}

Hence, smaller charge is -11.4 \times 10^{-5} and larger charge is 9.1 \times 10^{-5}.

8 0
3 years ago
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