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nasty-shy [4]
3 years ago
7

The short vertical parts adjacent to it also reach into the magnetic field and should experience forces. why can we neglect them

?
Physics
1 answer:
hammer [34]3 years ago
4 0

It is not only the horizontal part of the loop that dips into the magnetic field. We can neglect or disregard the horizontal parts of the loop that hollows into the magnetic fields since only the parts perpendicular to the magnetic field complement to it.

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The global winds and moisture belts indicate that large amounts of rainfall occur at
diamong [38]

it should be rising and converging

3 0
3 years ago
Explain how if a solid, liquid, and gas are put into individual containers how they fill the container.
Sav [38]

Answer:

serial in which container is filled

Solid -base of container

Liquid- above solid

Gas- above liquid

Explanation:

If any mixture of  matter in different state (that solid , liquid or gas )are kept in any container, then substance with higher density will be settled at lowest surface first and similarly the substance with lowest density will be at upper part of container.

In the given container we have to keep solid, liquid and gas

  • sold has the highest density,
  • gas the lowest density and
  • liquid has the density higher than gas but less than solid.

based on this

solid will be at surface of container

above sold will be liquid

above liquid will be presence of Gas

serial in which container is filled

Solid -base of container

Liquid- above solid

Gas- above liquid

8 0
4 years ago
Select all that apply. There MIGHT be more than one.
Stolb23 [73]
I believe the answer to your question is “Lithosphere plate boundaries”

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Hope this helps!:)
4 0
3 years ago
2. Why are the health-related fitness components more
ser-zykov [4K]

Answer:

Health-related physical fitness is primarily associated with disease prevention and functional health.

Explanation:

This is ur answer actually I just took it from Go ogle

7 0
3 years ago
Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.
kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, 4.67\times 10^7

(b) The number of radon nuclei have decayed by this time will be, 3.6\times 10^8

Explanation :

<u>For part (a) :</u>

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{3.82\text{ days}}

k=1.81\times 10^{-1}\text{ days}^{-1}

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.81\times 10^{-1}\text{ days}^{-1}

t = time passed by the sample  = 12 days

a = initially number of radon atoms  = 4.1\times 10^8

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}

a-x=4.67\times 10^7

Thus, the number of radon atoms will remain after 12 days is, 4.67\times 10^7

<u>For part (b) :</u>

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = (4.1\times 10^8)-(4.67\times 10^7)

The number of radon nuclei have decayed = 3.6\times 10^8

Thus, the number of radon nuclei have decayed by this time will be, 3.6\times 10^8

5 0
3 years ago
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