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kenny6666 [7]
3 years ago
12

I NEED THE RIGHT ANSWER ASAP NO LINKS !!! This is a Science question

Physics
2 answers:
Lunna [17]3 years ago
8 0

Answer:

Independent variable

Please mark brainliest and i'll be happy to help you again! :D

sergeinik [125]3 years ago
5 0

Answer:

independent variable

Explanation:

plz mark brainliest

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Kisachek [45]

Answer:

All points to the left of zero are negative

Explanation:

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2 years ago
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A radio has a 1.3 A current. If it has a resistance of 35 Ω, what is the potential difference?
DedPeter [7]

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22

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Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snork
fgiga [73]

Answer:

2354.4 Pa

40221 Pa

Explanation:

\rho = Density = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Depth

The pressure difference would be

\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa

The pressure difference in the first case is 2354.4 Pa

\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa

The pressure difference in the second case is 40221 Pa

7 0
3 years ago
A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a
Galina-37 [17]

Answer:

E   = (0.56 \times 10^8 ) r   \   \ N/c

Explanation:

Given that:

\rho_o = (10^{-3} ) \ c/m^3

R = (0.1) m

To find  the electric field for r < R by using Gauss Law

{\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)

For r < R

Q_{enclosed}=(\rho) ( \pi r^2 ) l

E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}

E= \dfrac{\rho ( r)}{2 \varepsilon_o}

where;

\varepsilon_o = 8.85 \times 10^{-12}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E   = (0.56 \times 10^8 ) r   \   \ N/c

4 0
3 years ago
Ram has power of 550 watt. What does it mean?
WARRIOR [948]
It means you can do 550 Newton Meters of work every second. Power is the rate of doing work, I hope this helps
4 0
3 years ago
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