Table is in the picture. and will mark brainstest.

O que levou a filosofia pensar em artes?

vredina [299]

Answer:

idk

Explanation:

4 0

3 years ago

When an atom is involved in a nuclear reaction:

nataly862011 [7]

I believe the answer would be , it may change from one element to another. I’m saying this because during nuclear transmutation is when a subatomic particle fired at the nucleus of an atom changes into a heavier element , to break the nucleus apart into two nuclei and energy.

4 0

3 years ago

HELP PLEASE!!!

Leya [2.2K]

For the first one 320

second

1200W

Data

R = 12 Ω ∆V = 120V I =? P =?

Solution:

According to Ohm’s law,

∆V = I R

I = ∆V / R

= 120 / 12

= 10 A

Power P = I ∆V

= 10 x 120

= 1200 W

Third

∆V = 120 V P = 60 W I =? R =?

Use the formula, P = I ∆V

I = P / ∆V = 60 / 120 = 0.5 A

∆V = I R

R = ∆V / I = 120 / 0.5 = 240 Ω

3 0

3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago

Whats the term for climate near equator

storchak [24]

The closer to the equator, the hotter the climate will be.

6 0

3 years ago

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