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3 years ago

What is the primary type of battery we use today to store energy?

2 answers:

4 0

Lithium-Ion batteries are commonly used in portable electronics and electric vehicles. These rechargeable batteries have two electrodes: one that's positively charged and contains lithium and another negative one that's typically made of graphite.

AleksAgata [21]3 years ago

3 0

Answer:

Lithium-Ion: Li-ion batteries are commonly used in portable electronics and electric vehicles. These rechargeable batteries have two electrodes: one that's positively charged and contains lithium and another negative one that's typically made of graphite.

Explanation:

Hope this helps :)

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Explain how atomic mass and molecular mass are determined

GaryK [48]

Molecular mass may be calculated by taking the atomic mass of each element present and multiplying it by the number of atoms of that element in the molecular formula. Then, the number of atoms of each element is added together. This value may be reported as a decimal number or as 16.043 Da or 16.043 amu.

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3 years ago

a block (mass 0.6kg) is released from rest at point A at the top of a ramp inclined at 36.9 degress above the horizontal. The bl

bija089 [108]

14.136 J as shown on the photo with two thought processes but overall same calculation

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3 years ago

A positively charged body exerts an electric field that causes a positively charged particle to move away from it. If work is do

ollegr [7]

Answer:

b. electric potential energy.

Explanation:

The energy required to move a charge against the electric field is known as the electric potential energy. As in above case positively charged body is exerting an electric field on the positive charge. As the same charges repel so the charge tend to move away. In order to push it towards the body we need a work done. As it is hard to push the positive charged particle towards the positive electric field. So in the cases like these particle occupies the electric potential energy.

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3 years ago

The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl

Aleks04 [339]

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the medium

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

$d=3.5\cdot 10^{-5}m$ is the separation between the plates

$A=0.063m \cdot 0.054m = 0.0034 m^2$ (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the capacitor is

$C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F$

Learn more about capacitors:

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#LearnwithBrainly

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3 years ago

A diagram of a closed circuit with a power source on the left labeled 6 V. There are 3 resistors in parallel, separate paths, co

AfilCa [17]

Answer:

Current: 1.0 Amperes

The minimum current is flowing through path D

Explanation:

We first find the equivalent resistance to the three resistors in parallel ( which is the total resistance of the circuit) via the equation:

$\frac{1}{R_e} =\frac{1}{R_B}+\frac{1}{R_C}+\frac{1}{R_D}\\\frac{1}{R_e} =\frac{1}{10}+\frac{1}{20}+\frac{1}{50}=0.17\\R_e=(1/0.17)\Omega\\R_e=5.88 \Omega$

with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

$V=I*R\\6V=I*5.88\Omega\\I=\frac{6}{5.88} Amp\\I=1.02A$

where the current comes in units of Amperes since all other the quantities are given in the SI system, and we can round this answer to 1.0 Amp following the request to round it to the tenth.

The current will be the lowest through the branch with the largest resistor due to the fact that less current will flow through the path of more resistance.

Than means that the lowest current will be registered through branch D where the 50 $\Omega$ resistor is.

8 0

3 years ago

Read 2 more answers