Ariana is accelerating her car at a rate of 4.6 m/s2 for 10 seconds. Her starting velocity was 0 m/s.

A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?

olchik [2.2K]

Answer:

$m=5.78\times 10^{-3}\ g$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Explanation:

Given:

quantity of charge transferred, $Q=31\ C$

No. of electrons in the given amount of charge:

As we have charge on one electron $1.602\times 10^{-19}\ C$

so,

$n_e=\frac{Q}{e}$

$n_e=\frac{31}{1.602\times 10^{-19}}$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Now if each water molecules donates one electron:

Then we require $n=1.935\times 10^{20}$ molecules.

Now the no. of moles in this many molecules:

$n_m=\frac{n}{N_A}$

where

$N_A=6.022\times 10^{23}$ Avogadro No.

$n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}$

$n_m=3.213\times 10^{-4}\ moles$

We have molecular mass of water as M=18 g/mol.

So, the mass of water in the obtained moles:

$n_m=\frac{m}{M}$

where:

m = mass in gram

$3.213\times 10^{-4}=\frac{m}{18}$

$m=5.78\times 10^{-3}\ g$

7 0

3 years ago

Reasons why using mirrors to generate electricity is not a common practice.

olganol [36]

A big part of the reason that mirrors are seldom if ever used to generate
electricity is the simple fact that there is no way to generate electricity using
mirrors. They are as useless for the purpose as smoke is, although there are
those who have used both items simultaneously to create the impression that
they have succeeded in that attempt.

7 0

3 years ago

Read 2 more answers

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

You can also describe position by using coordinates of _____and _______

SOVA2 [1]

Answer:

x and y

Explanation:

7 0

2 years ago

The standard measure used to compare sound intensities is the ______ .

uranmaximum [27]

The answer is Decibels.