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Temka [501]
4 years ago
13

What important factor does the ozone layer serve in earth's atmosphere?

Physics
1 answer:
Dmitry_Shevchenko [17]4 years ago
3 0

The Ozone layer is responsible for the absorption of harmful radiation from the sun before it hits the surface or exterior structure of our planet. It is known to be a belt of a known occurring gas, known as “ozone” which is sealed around the Earth’s atmosphere. It serves as a shield, which absorbs most of the sun’s ultraviolet radiation. 

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5.) How could we have improved this activity to be more precise?<br><br>ball on ramp ​​
AleksAgata [21]

Explanation:

I don't understand this question

could you please explain

7 0
3 years ago
two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will h
Aneli [31]

Answer:

On that line segment between the two charges, at approximately 0.7\; \rm m away from the smaller charge (the one with a magnitude of 5 \times 10^{-19}\; \rm C,) and approximately 1.3\; \rm m from the larger charge (the one with a magnitude of 20 \times 10^{-19}\; \rm C.)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let k denote the Coulomb constant, and let q denote the size of a point charge. At a distance of r away from the charge, the electric field due to this point charge will be:

\displaystyle E = \frac{k\, q}{r^2}.

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.  

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let q_1 = 5\times 10^{-19}\; \rm C and q_2 = 20 \times 10^{-19}\; \rm C. Assume that the electric field is zero at r meters to the right of the 5\times 10^{-19}\; \rm C point charge. That would be (2 - r) meters to the left of the 20 \times 10^{-19}\; \rm C point charge. (Since this point should be between the two point charges, 0 < r < 2.)

The electric field due to q_1 = 5\times 10^{-19}\; \rm C would have a magnitude of:

\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}.

The electric field due to q_2 = 20 \times 10^{-19}\; \rm C would have a magnitude of:

\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}.

Note that at all point in this section, the two electric fields E_1 and E_2 will be acting in opposite directions. At the point where the two electric fields balance each other precisely, | E_1 | = | E_2 |. That's where the actual electric field is zero.

| E_1 | = | E_2 | means that \displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}.

Simplify this expression and solve for r:

\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0.

\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0.

Either r = -2 or \displaystyle r = \frac{2}{3}\approx 0.67 will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, 0 < r < 2. Therefore, (-2) isn't a valid value for r in this context.

As a result, the electric field is zero at the point approximately 0.67\; \rm m away the 5\times 10^{-19}\; \rm C charge, and approximately 2 - 0.67 \approx 1.3\; \rm m away from the 20 \times 10^{-19}\; \rm C charge.

8 0
3 years ago
Please help! A net force of 2.0 N acts on a 2.0-kg object for 10 seconds. What is the object’s kinetic energy after that 10 seco
padilas [110]
For Newton's second law, the force F applied to the object of mass m will cause an acceleration a of the body:
F=ma
So, the acceleration is 
a= \frac{F}{m}= \frac{2.0 N}{2.0 kg}=1 m/s^2

The object undergoes through this acceleration for 10 seconds, t=10 s. Since it is an accelerated motion, we can find its final velocity after 10 seconds:
v_f = v_i + at=0 m/s+(1m/s^2)(10 s)=10 m/s
where v_i is the initial velocity of the object, which is zero since it starts from rest.

Finally we can calculate the final kinetic energy of the object, which is given by
K= \frac{1}{2}mv^2= \frac{1}{2}(2.0 kg)(10 m/s)^2=100 J
7 0
3 years ago
plz halp! For the PV graph shown above, explain whether heat is added to the gas or removed from the gas during the process indi
Delicious77 [7]

Answer:

There's no PV graph. kindly provide it so we help

5 0
3 years ago
Which isotope has 16 neutrons and is used to study nucleotides and nucleic acids?
qaws [65]
When naming isotopes, the number beside it indicates the mass number. Now mass number is the sum of protons and neutrons. 

We know that the mass number of a stable Sulfur atom is 32, so  we can rule out C as an isotope. The atomic number of Sulfur is 16 and that means that there are 16 protons. Now if we subtract 16 from 33, then that means there are 17 neutrons. So we can rule out D. 

Phosphorus on the other hand has an atomic number of 15. If we subtract 15 from 31 we will have 16. That means that Phosphorus-31 has 16 neutrons.

The answer would then be A. Phosphorus 31. 
7 0
4 years ago
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