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aniked [119]
3 years ago
9

12. A measuring cylinder has 150ml. of water in it. When a stone is immersed into the measuring cylinder, the water level raised

to 210cm. What is the volume of the stone?​
Physics
2 answers:
Serga [27]3 years ago
5 0
<h2>Correction: </h2>

The final level of water should be = 210

{cm}^{3}

or 210 ml

━━━━━━━━━━━━━

<h2>Given:</h2>
  • .Initial level of water = 150ml
  • Final level of water = 210 ml

━━━━━━━━━━━━━

<h2>Need to find:</h2>
  • Volume of stone =?

━━━━━━━━━━━━━

<h2>Solution:</h2>

The volume of stone will be equal to the change in level of water.

that's,

210 - 150 {cm}^{3}

= 60{cm}^{3}

Hence the <em>volume</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>stone</em><em> </em><em>is</em><em> </em><em>60</em><em> </em>{cm}^{3}

skelet666 [1.2K]3 years ago
4 0

Answer:

60ml

Explanation:

I'm going to assume you mean 210ml not centimeters. To find the volume all we do is subtract both values or with the formula [ f - i = v ]  where f = final amount and i = initial amount.

210 - 150 = 60ml

Best of Luck!

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ivanzaharov [21]

Answer:

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Explanation:

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For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

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Is energy matter? Is gravity matter? Why or why not
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A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

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a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

5 0
3 years ago
Read 2 more answers
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
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