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How do rechargeable batteries work? When a rechargeable battery is connected to electricity it changes the pressure inside and a

aalyn [17]

Question:

How do rechargeable batteries work?

Answer:

When a rechargeable battery is connected to electricity it changes the pressure inside and allows the rubber membrane to expand, thus releasing more energy next time its used. When plugged into the wall the electrons are sent back to rebuild the original dry cell (or metal). Rechargeable batteries are made exclusively of nitrogen which when exposed to electricity from and outlet can store a charge. When a battery is zapped by the electricity from the outlet it resets the system.

7 0

2 years ago

How many significant figures are in the number 17.95?

ANTONII [103]

Answer:

4

Explanation:

All of them are significant figures

6 0

4 years ago

12. Suppose a person uses a mechanical iack to lift half the weight of a car with a

Likurg_2 [28]

a) The force that must be applied is 73.5 N

b) The actual efficiency is 82 %

Explanation:

a)

Since the jack is 100% efficient, all the input work is converted into output work. So we can write:

$W_{in} = W_{out}\\F_{in} d_{in} = F_{out} d_{out}$

where:

$F_{in}$ is the input force applied on the jack

$d_{in}$ is the arm of the input force

$F_{out}$ is the output force applied on the jack

$d_{out}$ is the arm of the output force

Here we have:

$F_{out}=\frac{mg}{2}= \frac{(1200 kg)(9.8 m/s^2)}{2}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Solving for $F_{in}$ , we find the force that must be applied in input to lift the car:

$F_{in} = \frac{F_{out}d_{out}}{d_{in}}=\frac{(5880)(0.005)}{0.40}=73.5 N$

b)

The efficiency of the jack is given by the ratio between the output work and the input work:

$\eta = \frac{W_{out}}{W_{in}}=\frac{F_{out}d_{out}}{F_{in}d_{in}}$

where we have:

$F_{out}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Here we are told that the input force this time is

$F_{in}=90.0N$

Substituting into the equation, we find the new efficiency of the jack:

$\eta = \frac{(5880)(0.005)}{(90.0)(0.40)}=0.82$

Which means an efficiency of 82%.

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

8 0

4 years ago

A plane flies from base camp to lake a, 205 km away in the direction 20.0Â° north of east. after dropping off supplies it flies

miss Akunina [59]

Draw a diagram to illustrate the problem as shown in the figure below.

Camp A is 20° north of east from the camp, therefore
m∠CAB = 80°
where C => base camp.

Let d = distance from lake B to the base camp, at x° west of south.
Apply the Law of Cosines to determine d.
d² = 205² + 175² - 2*205*175*cos80⁰
= 6.0191 x 10⁴
d = 245.338 km

Apply the Law of Sines to obtain
$\frac{sin(x+30)}{205}= \frac{sin80^{o}}{245.338}\\ \\ sin(x+30)=( \frac{205}{245.338})sin80^{o}= 0.8229$
x+30 = sin⁻¹ 0.8229 = 55.4°
x = 25.4°

Answer:
The distance from lake B to base camp is 243.3km (nearest tenth).
The direction is 25.4° west of south.

7 0

3 years ago

Kinetic energy into electrical energy

jeyben [28]

<span>A wind turbine is a machine for converting the kinetic energy in wind into electrical energy. Moving air (wind) is used to rotate a machine called a turbine. Inside the wind turbine is an electrical generator. It's the electricalgenerator that transforms the kinetic energy into electrical energy.</span>

3 0

4 years ago

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