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3 years ago

What are limitations of the calculated and indirect volume measurement?

Physics

2 answers:

PilotLPTM [1.2K]3 years ago

7 0

Answer: Not suitable for large volume and irregular shaped object.

Explanation: The answer refers to a high school physics experiment where candidates are required to determine the density of an object which is regularly or irregularly shaped.

Therefore, an object whose shape is irregular although its volume might be as same as a regular shape i.e, (cuboid), might be difficult to fit in the container provided for the density.

tresset_1 [31]3 years ago

3 0

Answer: The calculated volume measures the side of the cube to get the volume. An indirect measurement of volume measures the displacement of a liquid by a solid.

Explanation:

Volume is the amount of the space occupied by an object. Volume can be used to measure the amount of the space occupied by the solid, liquid and gas.

When the direct measurement is not possible then the indirect measurement is used by taking the small part that object to calculate the whole part of that.

The limitations of the calculated and indirect volume measurement are as follows,

The calculated volume measures the side of the cube to get the volume. An indirect measurement of volume measure the displacement of a liquid by a solid. For example, the volume of the liquid is displaced by the sponge is not equal to the volume of the sponge.

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Katyanochek1 [597]

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8 0

3 years ago

The Earth's electric field creates a potential that increases 100 V for every meter of altitude. If an object of charge 4.5 mC a

castortr0y [4]

Answer:

The final kinetic energy is $K = 1.1 \ J$

Explanation:

From the question we are told that

The electric field is $E = 100 \ V/m$

The charge on the object is $q = 4.5 mC = 4.5 *10^{-3} \ C$

The mass of the object is $m_o = 68 \ g = 0.68 \ kg$

The distance moved by the object is $d = 1.0 \ m$

The workdone on the object by the fields is mathematically represented as

$W = [qE + mg]d$

Now this workdone is equivalent to the final kinetic energy so

$K = W = [qE + mg]d$

substituting values

$K = W = [4.5*10^{-3 } *100 + 0.68 * 9.8]* 1$

$K = 1.1 \ J$

8 0

3 years ago

Determine the magnitude of the current flowing through a 4.7 kilo ohms resistor if the voltage across it is (a) 1mV (b) 10 V (c)

mariarad [96]

Answer:

213 nA

2.13 mA

851e^-t μA

Explanation:

We have a pretty straightforward question here.

Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as

V = IR, since we need I, we can write that

I = V/R

a) at V = 1 mV

I = (1 * 10^-3) / 4.7 * 10^3

I = 2.13 * 10^-7 A or 213 nA

b) at V = 10 V

I = 10 / 4.7 * 10^3

I = 0.00213 A or 2.13 mA

c) at V = 4e^-t

I = 4e^-t / 4.7 * 10^3

I = 0.000851e^-t A or 851e^-t μA

5 0

2 years ago

WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of

nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

$A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km$