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nydimaria [60]
3 years ago
7

Manganese dioxide is used in the dry cell as​

Physics
1 answer:
slamgirl [31]3 years ago
5 0

Answer:

Electrolytic manganese dioxide (EMD) is used in zinc-carbon batteries together with zinc chloride and ammonium chloride. EMD is commonly used in zinc manganese dioxide rechargeable alkaline (Zn RAM) cells also. For these applications, purity is extremely important.

Explanation:

hope this helps...

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kykrilka [37]
To keep cars safe...............
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3 years ago
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Let's begin by determining the equilibrium position of a seesaw pivot. You and a friend play on a seesaw.Your mass is____
expeople1 [14]

Answer:

1.2 m from the left end

Explanation:

M = mass of the person = 90 kg

m  = mass of friend = 60 kg

r  = distance of pivot from the person or from the left end

L  = length of the seesaw board = 3.0 m

s  = distance of pivot from the friend or from right end = L - r = 3 - r

Using equilibrium of torque about the pivot

M g r = m g s\\(90) r = (60) (3 - r)\\90 r = 180 - 60 r\\90 r + 60 r = 180 \\150 r = 180 \\r = 1.2

6 0
3 years ago
A standing wave of the third overtone is induced in a stopped pipe, 3 m long. The speed of sound is 340 m/s. The number of antin
Leokris [45]

Answer:

overtone- one over the first

n skips by twos

4 antinodes

500 Hz

Explanation: Hope this helps :)

7 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Studentka2010 [4]
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Speed = distance / time

Substitute in your values:

Speed = 720 / 8

Speed = 90

They traveled 90 miles per hour

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

8 0
4 years ago
Read 2 more answers
4000 J of heat energy is applied to a 70 g sample of water initially at 35oC. What is the final temperature of the sample?
BARSIC [14]

Answer:

48.7°C

Explanation:

Step one:

given data

Quantity of heat = 4000J

mass of sample= 70g= 0.07kg

initial temperature T1=35°C

Water has a specific heat capacity of 4182 J/kg°C

Required

The final temperature T2

Step two:

Q= mcΔT

4000= 0.07*4182(T2-35)

4000=292.74(T2-35)

4000=292.74T2-10245.9

collect like terms

4000+10245.9=292.74T2

14245.9= 292.74T2

divide both sides by 292.74

T2= 14245.9/ 292.74

T2=48.7°C

8 0
3 years ago
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