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2 years ago

What is the difference between Daniel and Leclanché cell

Physics

1 answer:

Thepotemich [5.8K]2 years ago

3 0

Explanation:

Leclanche cell : Leclanche cell is also known as dry cells (fig 2). It contains an electrolyte of ammonium chloride. Carbon acts as a cathode while zinc is an anode.

Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/959247-difference-between-daniel-and-leclanch-cells.htmlDaniel cells: It is a primary cell. It consists of a zinc rod and a copper vessel (in fig 1). It can be seen that a zinc rod is put in dilute sulfuric acid which is contained in a porous pot. Daniel cell works on the principle of the redox reaction. The electrons pass form zinc rod to wire. Then it passes to the copper rod. Thus, a current is produced.

Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/959247-difference-between-daniel-and-leclanch-cells.html

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How has modern safety equipment found in automobiles helped to counteract Newton’s First Law of Motion?

Charra [1.4K]

Answer:

Well. the law says "an object in motion stays in motion" So seatbelts could be one. they stop us humans from continuing to move when the car stops.

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3 years ago

A car travels 3 km north, then turns and travels 4 km west. What is the car's displacement?

Olin [163]

Explanation:

Displacement = Shortest path Travelled from starting point to final point..

Displacement formula = Final position - Initial position

= 7km - 3km

= 4km

^Displacement is 4km

If this answer helps you plz mark as brainlist..

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3 years ago

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Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

-Dominant- [34]

Answer:

The required work done is $6.5\times10^{9}\ J$

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

$U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}$

Put the value into the formula

$U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}$

$U_{A}=-3.44\times10^{10}\ J$

We need to calculate the potential energy

Using formula of potential

$U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}$

Put the value into the formula

$U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}$

$U_{B}=-2.14\times10^{10}\ J$

We need to calculate the value of $k_{A}$

Using formula of $k_{A}$

$k_{A}=-\dfrac{1}{2}U_{A}$

Put the value into the formula

$k_{A}=\dfrac{1}{2}\times3.44\times10^{10}$

$k_{A}=1.72\times10^{10}\ J$

We need to calculate the value of $k_{B}$

Using formula of $k_{B}$

$k_{B}=-\dfrac{1}{2}U_{B}$

Put the value into the formula

$k_{B}=\dfrac{1}{2}\times2.14\times10^{10}$

$k_{B}=1.07\times10^{10}\ J$

We need to calculate the work done

Using formula of work done

$W=\Delta K+\Delta U$

$W=(k_{B}-k_{A})+(U_{B}-U_{A})$

$W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})$

$W=\dfrac{1}{2}(U_{B}-U_{A})$

Put the value into the formula

$W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})$

$W=6.5\times10^{9}\ J$

Hence, The required work done is $6.5\times10^{9}\ J$

7 0

2 years ago

A 4-kg mass moving with speed 2 m/s, and a 2-kg mass moving with a speed of 4 m/s, are gliding over a horizontal frictionless su

Akimi4 [234]

Answer:

B) The 2-kg mass travels twice as far as the 4-kg mass before stopping

Explanation:

As we know that both mass have same horizontal force opposite to their motion

So we will have

$a = \frac{F}{m}$

$a_1 = \frac{F}{4}$

$a_2 = \frac{F}{2}$

now the stopping distance of an object moving with initial speed v is given as

$v_f^2 - v_i^2 = 2(-a) d$

$d = \frac{v^2}{2a}$

so here we have

$d_1 = \frac{2^2}{\frac{F}{4}}$

$d_1 = \frac{16}{F}$

for other object we have

$d_2 = \frac{4^2}{\frac{F}{2}}$

$d_2 = \frac{32}{F}$

So correct answer will be

B) The 2-kg mass travels twice as far as the 4-kg mass before stopping

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3 years ago

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

Greeley [361]

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3 years ago

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