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Genrish500 [490]
3 years ago
9

Regular exercise helps to options: lower blood pressure

Physics
2 answers:
marysya [2.9K]3 years ago
6 0
The answer is all of the above
disa [49]3 years ago
4 0
All of dee above is da answer ku
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How can a magnetic field be produced, using a wire, a battery, and and a nail?
Aleks [24]
It is fairly easy to build an electromagnet. All you need to do is wrap some insulated copper wire around an iron core. If you attach a battery to the wire, an electric current will begin to flow and the iron core will become magnetized. When the battery is disconnected, the iron core will lose its magnetism. Follow these steps.
Step 1 - Gather the Materials
One iron nail fifteen centimeters (6 in) long
Three meters (10 ft) of 22 gauge insulated, stranded copper wire
One or more D-cell batteries
Step 2 - Remove some Insulation
Step 3 - Wrap the Wire Around the Nail
Step 4 - Connect the Battery
5 0
3 years ago
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Help please fast + i dont want an answer from g**gle .
velikii [3]
7) 6, i believe, (Cu) 1 atom+ (S) 1 atom+(0)4 atoms.
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What happens to a decimal place in the number when you are converting from a smaller metric unit to a larger unit?
Natasha_Volkova [10]

I dont have any idea :P

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3 years ago
The Earth's gravity keeps us from flying out to space. How can you explain that we have a stronger gravitational attraction to t
julia-pushkina [17]
It's the fourth choice.

This is because, since we are closer to the Earth, the Earth will have a stronger gravitational pull on us since again, we are closer.
That also explains tides, but that's just getting off topic. Hope I helped.
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3 years ago
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An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
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