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julia-pushkina [17]

3 years ago

5

How would you arrange the objects below from the least to greatest volume?

1 answer:

EastWind [94]3 years ago

5 0

Answer:

Empty bottle, Balloon, Waterbottle Full Then rock

Explanation:

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What is an Amplitude

dangina [55]

Amplitude is a measurement of the magnitude of displacement (or maximum disturbance) of a medium from its resting state, as diagramed in the peak deviation example below (it can also be a measurement of an electrical signal's increased or decreased strength above or below a nominal state).

3 0

3 years ago

Read 2 more answers

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

3 years ago

15. Find the speed of a disc of radius 0.5 meters and mass 2-kg at the base of the incline. The disc starts at rest and rolls do

mote1985 [20]

Explanation:

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7 0

3 years ago

Work is the ability to cause what?

madreJ [45]

Work is the ability to cause life change??

8 0

3 years ago

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A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp

Sergeu [11.5K]

Answer:

$U=1.29\times 10^{-7}\ J$

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm ,K₁= 4.7

d₂=2.1 mm , K₂=2.6

We know that capacitance given as

$C_1=\dfrac{K_1\varepsilon _oA}{d_1}$

$C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_1=1.26\times 10^{-10}\ F$

$C_2=\dfrac{K_2\varepsilon _oA}{d_2}$

$C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_2=0.701\times 10^{-10}\ F$

Net capacitance

$C=\dfrac{C_1C_2}{C_1+C_2}$

$C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F$

$C=4.5\times 10^{-11}\ F$

We know that stored energy given as

$U=\dfrac{CV^2}{2}$

V= 76 V

$U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J$

$U=1.29\times 10^{-7}\ J$

3 0

3 years ago