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bekas [8.4K]
3 years ago
11

How much energy is given to each coulomb of charge passing through a 6 V battery

Physics
1 answer:
ELEN [110]3 years ago
4 0
1 volt = 1 joule per coulomb. Current doesn't actually pass 'through' a battery. But if it did, then each coulomb would gain or lose 6 joules in traversing 6 volts, depending on its sign, and whether it climbed or fell.
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How long does it take for a 4 inch candle to burn out
kotykmax [81]

What does "a 4-inch candle" mean ?
Is that the length of the candle or its diameter ?

The life of a candle depends on its length, its diameter, the
thickness and starting-length of its wick, the exact substance
of which it's composed ... the type of wax ... and the motions
of the air around it.

I'll always remember the candle shop I visited ... I don't remember
what I was doing in a candle shop, but I was there.  There was a
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8 0
3 years ago
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In a simple circuit, if you replace one resistor with a resistor of a higher value will the reading the ammeter increase, decrea
Citrus2011 [14]
The ammeter reading (current) will decrease.
V=IR
I = V/R
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6 0
4 years ago
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A hiker walks 9.4 miles at an angle of 60° south of west. Find the west and south components of the walk
Allisa [31]

Answer:

Ax= -4.7 miles

Ay= -8.1

Explanation:

3 0
4 years ago
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
4 years ago
A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational
Naya [18.7K]

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

4 0
3 years ago
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