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larisa86 [58]
3 years ago
7

HELP NEEDED

Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

D the development of an ozone layer

Explanation:

the rest don't relate to the atmosphere

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Which wave has a longer period and how many times, the wave with a frequency of 7000Hz or the wave with a frequency of 21.000Hz?
Nuetrik [128]

Answer:

I would say the answer is the wave of 21.000Hz

Explanation:

Because it has more frequency, and as more frequency you add, the time or longer period also increases.

3 0
2 years ago
For a reaction to occur what must happen to the energy in order to break the chemical bond
Andrej [43]
Energy needs to realease
6 0
3 years ago
How do you tell a stars life expectancy based on an hr diagram?
e-lub [12.9K]
The X-axis of the H-R Diagram indicates the star's surface temperature in degrees Kelvin.  The Y-axis, on the other hand, indicates luminosity, or brightness.

Main sequence refers to a roughly diagonal, slightly S-curved line stretching between the upper-left and lower-right corners on which main sequence stars chart. They maintain a predictable relationship between luminosity and temperature: the brighter, the hotter. The upper-right quadrant of the H-R diagram is home to newly discovered red giants while the lower-left quadrant of the H-R Diagram belongs almost exclusively to white dwarfs.
8 0
3 years ago
A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c
tatuchka [14]

We have that the values for F north, F east, F up are

  • F_N=1.09090909*10^{-5}
  • F_E=5.18181818*10^{-6}
  • F_E=2*10^{-6}

From the Question we are told that

electric force F_1 = 1.2 x 10^{-3} N(N)

electric force , F_2=5.7 x 10^{-4} N(E)

electric force , F_3=2.2 x 10^{-4} N (U)

charge on this ball one q_1= 110 nC.

charge on this ball two q_2= -50 nC.

Generally the equation for the F north  is mathematically given as

F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}

F_N=1.09090909*10^{-5}

For F East

F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}

F_E=5.18181818*10^{-6}

For F UP

F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}

F_E=2*10^{-6}

For more information on this visit

brainly.com/question/21811998

5 0
2 years ago
A ball rolls with a speed of 2.0m/s across a level table that is 1.0m above the floor. Upon reaching the edge of the table it fo
Mekhanik [1.2K]

Answer:0.58 m

Explanation:

The initial velocity of the ball is u = 2.0 m/s

The height of the table is, h = 1.0 m

The ball falls in vertical direction under acceleration due to gravity.

Time taken for ball to hit the floor:

h= ut + 0.5gt² ( from the equation of motion)

1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²

Solving this for t,

t = 0.29 s ( we have neglected the negative value of t)

In the same time, the ball would cover a horizontal distance of :

s = u t

⇒s = 2.0 m/s×0.29 s = 0.58 m

Thus, the landing spot is 0.58 m away from the table.

6 0
3 years ago
Read 2 more answers
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