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3 years ago

9

All battery powered devices, run off of what type of power?

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

Electrical energy

Explanation:

electrical energy Essentials. A battery is a device that stores chemical energy and converts it to electrical energy. The chemical reactions in a battery involve the flow of electrons from one material (electrode) to another, through an external circuit. The flow of electrons provides an electric current that can be used to do work.

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When an electron de-excites from the third quantum level to the second, and then to the ground state, two photons are emitted. T

Leokris [45]

Answer:

ground state

Explanation:

Lets take

n=3 ,n=2 ,n=1 are the energy level.

Energy level n=1 is the ground energy level.

The energy from 3 to 1 = hν

The energy from 3 to 2 = hν₁

The energy from 2 to 1 = hν₂

We can say that

hν = hν₁ + hν₂

If the electron were de-excitation from the third level to ground level then the sum of emitted frequency will be equal to the frequency of a single electron.

Therefore the answer is ground state.

3 0

3 years ago

What will occur when the trough of wave A overlaps the trough of wave B?

densk [106]

Constructive interference will occur, which means the waves will combine.

In destructive inference, the waves cancel each other out.

Hope this helps :)

3 0

3 years ago

Read 2 more answers

The density for gold is 19.3 g/cm3. What would be the mass of a 45 cm3 piece of gold?

lyudmila [28]

Answer:

868.5 g

Explanation:

Mass= Density x Volume

Mass= 19.3 x 45

=868.5

6 0

3 years ago

Read 2 more answers

A locomotive is pulling 8 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (wi

nikitadnepr [17]

Answer:106560 N

Explanation:

Given

Let Tension between 2 and 3 car be $T_{23}$ and between 3 &4 is $T_{34}$

$T_{23}-T_{34}=ma$

mass of freight car =37,000 kg

acceleration of car $=0.48 m/s^2$

$T_{34}$ is accelerating all freights behind 3

therefore

$T_{34}=5\times ma$

$T_{34}=5\times 37000\times 0.48=88,800 N$

Thus

$T_{23}=T_{34}+ma$

$T_{23}=88,800+37000\times 0.48=88,800+17,760=106560 N$

4 0

4 years ago

X-rays with an energy of 301 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 77.5^{\circ}

Alex73 [517]

Answer:

$6.03\cdot 10^{-12} m$

Explanation:

First of all, we need to find the initial wavelength of the photon.

We know that its energy is

$E=301 keV = 4.82\cdot 10^{-14}J$

So its wavelength is given by:

$\lambda = \frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.82\cdot 10^{-14} J}=4.13\cdot 10^{-12}m$

The formula for the Compton scattering is:

$\lambda' = \lambda +\frac{h}{mc}(1-cos \theta)$

where

$\lambda$ is the original wavelength

h is the Planck constant

m is the electron mass

c is the speed of light

$\theta=77.5^{\circ}$ is the angle of the scattered photon

Substituting, we find

$\lambda' = 4.13\cdot 10^{-12} m +\frac{6.63\cdot 10^{-34} Js)}{(9.11\cdot 10^{-31}kg)(3\cdot 10^8 m/s)}(1-cos 77.5^{\circ})=6.03\cdot 10^{-12} m$

7 0

3 years ago