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2 years ago

Which solid state component allows a mobile phone to store data even if the battery is removed

Physics

1 answer:

Talja [164]2 years ago

6 0

Answer:

Capacitor

Explanation:

Capacitor ; A capacitor is a passive two-terminal electrical component that can store energy in an electric field electrostatically. It works as a small rechargeable battery that stores electricity. However, unlike a battery, it can charge and discharge in the split of a second. Capacitors are widely used to build different types of electronic circuits.

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Convert 27,549 into scientific notation

dezoksy [38]

2.7549 x 10^4 is the answer I hope this helped u

7 0

2 years ago

How do you find acceleration due to gravity with time and height given?

Feliz [49]

Here, height is given which will be the distance for a freely falling object.

The velocity will be

$v=\text{ }\frac{h}{t}$

and the acceleration will be

$a=\frac{v}{t}$

In this way, the formula works.

3 0

8 months ago

A carriage of 20 kg is pulled with a force of 35 N. How far the carriage will go

Gennadij [26K]

Answer:

2.71 m

Explanation:

Force is the product of mass and acceleration

F=m*a

Work done is the product of force and distance

Work done=F*d

In this case;

F= 35 N

Work done = 95 J

95 =35 * d

95 /35 = d

2.71 m= d

6 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

6 0

3 years ago

If the speedometer of your car reads a constant speedometer of your car reads a constant speed of 40km/hr , can you say the car

Harrizon [31]

I think u can say thats a constant velocity, but remember if ur turning, or going around a curve, that is also changing velocity. Hope this helps have a great day!

4 0

3 years ago