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3 years ago

Which elements do hydrogen fuel cells combine to produce electricty

1 answer:

3 0

A fuel cell combines hydrogen and oxygen to produce electricity, heat, and water. Fuel cells are often compared to batteries. Both convert the energy produced by a chemical reaction into usable electric power.

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The resistance of resistor is greater for:

Monica [59]

Answer:

c: long and thin resistor.

Explanation:

The resistance of a resistor is given by:

R = ρ*L/A

where:

R = resistance

ρ = resistivity (depends on the material)

L = length of the material

A = cross-sectional area of the material

We can see that the length is on the numerator, which means that if we increase the length, then the resistance is increased.

We also can see that the cross-sectional area is on the denominator, then if we increase the area (for example, with a ticker resistor) the resistance decreases.

Then if we want to maximize the resistance, we need to have a long and thin resistor, so the correct answer is c.

8 0

3 years ago

does anyone know how to do this? the scenario is ""dominique reads that race cars have wide tires because the increased area of

Usimov [2.4K]

Answer:

Stupid

Explanation:

Because there is never a answer when we are trying to find one

7 0

3 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

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

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

3 years ago

Capacitances of 10uF and 20uF are connected in parallel,

jarptica [38.1K]

Answer:

The equivalent capacitance will be $15\mu F$

Explanation:

We have given two capacitance $C_=10\mu F\ and\ C_2=20\mu F$

They are connected in parallel

So equivalent capacitance $C=C_1+C_2=10+20=30\mu F$

This equivalent capacitance is now connected in series with $30\mu F$

In series combination of capacitors the equivalent capacitance is given by $\frac{1}{C}=\frac{1}{30}+\frac{1}{30}$

$C=\frac{30}{2}=15\mu F$

So the equivalent capacitance will be $15\mu F$

5 0

3 years ago

Neglecting air resistance, if a package is dropped from an airplane flying 100 km/hr at an altitude of 200 meters, how far from

UkoKoshka [18]

Answer:

178 m

Explanation:

4 0

3 years ago