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Andre45 [30]
2 years ago
5

Please Please Please Can Help Me On This Question!!!!! I Give Thanks!!!!

Physics
1 answer:
Kaylis [27]2 years ago
5 0
I'm doing questions in grade 10 math working on imperial system how should I give my answers in decimals or fraction it's an adult learning experience
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An electric iron operating at 120 volts draws 10. amperes of current. how much heat energy
faltersainse [42]
The complete question is how much heat energy is delivered by the iron in 30 seconds.
Heat is given by IVt, Where I is the current in Amperes, V is the voltage, and t is the time in seconds,
Therefore; 
Heat energy = 120 × 10 ×30
                     = 36000 joules or 36 kJ
Heat energy is measured in joules.
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3 years ago
The water is reflecting light, Is this specular or diffuse reflection? explain your answer​
FinnZ [79.3K]
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A 15-watt bulb is connected to a circuit that has a total of 60. Ω of resistance. How many electrons are passing through that bu
Mariulka [41]

Answer:

3.2075*10^16

Explanation:

Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.

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2 years ago
Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

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3 years ago
What are the risk of obesity in middle adulthood
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