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3 years ago

Batteries don’t run forever. Why do you think batteries run down?

Engineering

2 answers:

Marina86 [1]3 years ago

3 0

Because there is no such thing as immortal batteries.

When a battery is connected to a circuit, the charge moves through the circuit, and a chemical reaction occurs inside that separate the charges. The strength of this reaction diminishes over time and the battery eventually dies.

igomit [66]3 years ago

3 0

Answer;

Each charge cycle degrades the electrodes just a little bit more, meaning the battery loses performance over time, which is why even rechargeable batteries don't keep on working forever. Over the course of several charge and discharge cycles, the shape of the battery's crystals becomes less ordered.

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Examine a process whereby air at 300 K, 100 kPa is compressed in a piston/cylinder arrangement to 600 kPa. Assume the process is

professor190 [17]

Answer:

See attachment and explanation.

Explanation:

- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.

- The plot of the graph is given in attachment.

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3 years ago

X+3=2 x=?? No spamming

PtichkaEL [24]

Answer:

x+3=2

x=2-3꧁

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3 years ago

Can you help me with this

Fiesta28 [93]

the function is to provide sealed combustion so that the loss of gas is minimized

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2 years ago

You want to find all files on your server that have either the SGID or SUID permission set. Which command should you use to obta

aalyn [17]

Answer:

For SGID you type this

$ find . -perm /4000

For SUID you type this

$ find . -perm /2000

Explanation:

Auxiliary file permissions, that are commonly referred to as “special permissions” in Linux are needed in order to easily find files which have SUID (Setuid) and SGID (Setgid) set.

After typing

$ find directory -perm /permissions

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3 years ago

A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a

vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60 .................1

put here value

15 × $10^{3}$ = 2π×1750×T / 60

T = 81.84 Nm

and

torsion = T / Z ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³ .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

torsion = σy / factor of safety

416.75 / d³ = 530 × $10^{6}$ / 3

so d = 0.0133 m

so diameter is 14 mm

3 0

3 years ago