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Mazyrski [523]
4 years ago
10

What substance is reduced in a lead storage battery?

Physics
1 answer:
Norma-Jean [14]4 years ago
3 0

In lead acid battery, chemical reactions are induced to do work on charge and produce a voltage difference between the two output terminals.

In a lead storage battery, the negative terminal is composed of lead electrode (Pb2+) while in the positive terminal it is composed of lead dioxide electrode (PbO2).

The chemical half reactions in each electrode are:

<span>Negative Terminal: Pb2+  +  SO4(2-) ---> PbSO4 + 2e</span>

<span>            Lead electrode is oxidized to supply electrons.</span>

Positive Terminal: PbO2 + H2SO4 + e ---> PbSO4

<span>            Lead dioxide electrode is reduced by the addition of electrons.</span>

 

<span>Answer: Lead dioxide is reduced.</span>

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From the top of a cliff, a person tosses a pebble straight downward with an initial velocity of -9.0 meters/second. After 0.50 s
Irina-Kira [14]
Y - yo = Vo*t - g * (t^2) / 2

Vo = - 9.0 m/s
t = 0.50 s

=> y - yo = -9.0 m/s * 0.5 s - 9.8 m/s^2 * (0.5s)^2 / 2 = - 4.5m - 1.225m = - 5.725 m.

Answer: option c) - 5.7
8 0
4 years ago
Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s,
SSSSS [86.1K]

Answer:

Time period between the successive beats will be 0.1703 sec

Explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano A\lambda _A=0.766m

Wavelength of piano  B\lambda _B=0.776m

So frequency of piano A f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz

Frequency of piano B f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec

So time period between the successive beats will be 0.1703 sec

4 0
4 years ago
A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm
8_murik_8 [283]

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position x_{i}=5.00\ cm

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx

Put the value into the formula

\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}

v^2=\dfrac{2\times12.3285}{1.00}

v^2=24.657

v=4.96\ m/s

Hence, The speed of the block is 4.96 m/s.

6 0
3 years ago
Please help me asdrtyuio
marta [7]

Answer:

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Part 1: Fill in the SI unit for each of the following measurements. 1. Time: 2. Length: 3. Mass: 4. Temperature:
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Answer:

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...

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