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3 years ago

How many drive cycles can be completed with a initially fully charged battery?

Physics

1 answer:

IRINA_888 [86]3 years ago

5 0

Answer:

600 cycles

A phone battery is typically designed to last around 500 to 600 cycles, and a cycle is defined as charging a completely dead battery to 100% then draining it to zero again.

Explanation:

I hope it can help

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A spinning top spinning at 16rad/s takes 88.9s to come to a complete stop. Find the angular acceleration of the top.

AURORKA [14]

Answer:

The angular acceleration of the top is 0.18 rad/s²

Explanation:

Given;

angular velocity of the top, ω = 16 rad/s

time it takes to come to a complete stop, t = 88.9s

The angular acceleration of the top, is calculated as;

$\alpha = \frac{\omega}{t}$

α is the angular acceleration

ω is angular acceleration

t is the time

$\alpha = \frac{\omega}{t}\\\\\alpha = \frac{16}{88.9}\\\\\alpha =0.18 \ rad/s^2$

Therefore, the angular acceleration of the top is 0.18 rad/s²

5 0

3 years ago

What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.

kirill [66]

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, $q = 5.7\ \mu C=5.7\times 10^{-6}\ C$

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

$E=\dfrac{q}{A\epsilon_o}$

$A=\dfrac{q}{E\epsilon_o}$

$A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}$

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.

8 0

3 years ago

lab safety test Safety goggles and an apron must be worn when handling chemicals labeled corrosive because they:________.

Sunny_sXe [5.5K]

Answer:

The terms of biosafety in a laboratory require that the person must not have the eyes and skin exposed.

Explanation:

A corrosive substance is a substance that can damage a surface when they come into contact.

These substances represent a danger in people since they can burn the eyes, the skin, and the inside of the body since the inhalation of gases can burn the respiratory tract.

These chemical burns can be avoided by properly following the biosafety protocol that a laboratory requires the use of masks, goggles, gloves, and an apron or lab coat.

5 0

4 years ago

Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display

goldfiish [28.3K]

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ .

This can be written as $\lambda=\frac{h}{p}$ .....(1) where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by $p=\frac{E}{c}$ ...(2) where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

$p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}$

(b) As we know that kinetic energy is given by

$K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p$ ...(3)

Where mass of electron is $9.1\times10^-^3^1 kg$ .

Putting K.E = 3eV in equation (3) , we get

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}$

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m$