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rosijanka [135]
3 years ago
5

A water bath in a physical chemistry lab is 1.55 m long, 0.710 m wide, and 0.570 m deep (high). If it is filled to within 3.55 i

n from the top, how many liters of water are in it?
Physics
1 answer:
Lesechka [4]3 years ago
7 0

Answer:

528 liter.

Explanation:

Volume of the tank(cuboid) = l*b*h

But volume of the water = l*b*h

Where

l= length of the tank

b = width of the tank

h = the length from the bottom of the tank,

3.55 in to m,

0.09017m

Length of the water in the tank = 0.570 - 0.09017

= 0.47983 m.

Volume = 0.47983*0.710*1.55

= 0.528 m3.

1 m3 = 1000 liter.

0.528 m3 = 0.528*1000

= 528 liter

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Answer:

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Explanation:

i think

5 0
2 years ago
In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
sweet-ann [11.9K]

Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

5 0
3 years ago
the angular speed of an automobile engine is increased at a constant rate from 1300rev/min to 2000rev/min in 3s (a) what is its
GalinKa [24]

Answer:

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Explanation:

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8 0
3 years ago
Can lamp that works on a 2.5 v work on a 1.12 v ?​
12345 [234]

Answer:

Explanation:

Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.

5 0
3 years ago
The density of mobile electrons in copper metal is 8.4 × 1028 m-3. Suppose that i= 4.4 × 1018 electrons/s are drifting through a
neonofarm [45]

Answer: 405.3 minutes

Explanation: In order to explain this problem we have to use the following:

Fisrtly we calculate the volume of the wire, this is given by:

Vwire=π*r^2*L where r and L are the radius and L the length of teh wire, respectively.

Vwire=π*1.25*10^-3*0.26=1.27*10^-6 m^3

then the number of the total electrons in tthe wire volume is given by;

n° electrons in the wire=ρ*Vwire=8.4*10^28*1.27*10^-6 m^3=1.07 *10^23

Finally, considering the current in the wire equal to 4.4*10^18 electrons/s

the time consuming to extract all the electrons from the wire is given by:

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equivalent to 405.3 minutes

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3 years ago
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