How to charge a laptop battery without a charger?

2 answers:

7 0

I dont know if it right but <span>Use a universal power adapter. This is perhaps the most obvious solution to your battery woes. Readily found at most retail outlets that carry electronics, a universal power adapter can range anywhere in price from $30 to $100 or more. The adapter comes with multiple tips, one of which will likely fit your laptop’s charging port. When plugged in, the adapter will not only power your laptop, but will charge its battery as well.</span>

WITCHER [35]2 years ago

5 0

Power Adapter will help u a lot

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Which statement is true? CTP hydrolysis can be a source of chemical energy. dATP is the most common source of chemical energy in

Gemiola [76]

1. Statement A is true.

When GTP is hydrolysed, the free energy of hydrolyses is used to power or drive reactions that are favourable energetically.

2. Statement B it true.

ATP is a complex chemical that gives energy for the activities in many living cells. During hydrolyses, chemical energy stored in the energy-rich phosphoanhydride is released. Hence its a common source of chemical energy in cells.

3. Statement C is false.

The hydrolyses of ATP to ADP in the presence of phosphate, releases one mole of ATP which is estimated to be -57Kj/mol not 14Kj/mol. Below is the equation;

ATP + H20 -----> ADP + Pi + Free energy.

4. Statement D is false.

GTP stands for guanosine triphosphate.

ATP stands for adenosine triphosphate

ADP stands for adenosine diphosphate.

3 0

3 years ago

Near the center of our galaxy, hydrogen gas is moving directly away from us in its orbit about a black hole. The electromagnetic

klemol [59]

To solve this problem it is necessary to apply the concept related to wavelength, specifically when the wavelength is observed from a source that is in motion to the observer.

By definition the wavelength is given defined by,

$\lambda_{obs} = \lambda_s \sqrt{\frac{1+u/c}{1-u/c}}$

Where

$\lambda_{obs}$ = Observed wavelength

$\lambda_s$ = Wavelength of the source

c = Speed of light in vacuum

u = Relative velocity of the source to the observer

According to our data we have that the wavelength emitted from the galaxy is 1875nm which is equal to the wavelength from the source, while the wavelength from the observer is $\lambda_{obs}=1945nm$