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4 years ago

Objects: A balloon full of helium in the air. Air- 1.27 kg/ml or Helium Balloon- 0.33 kg/ml

Physics

1 answer:

dem82 [27]4 years ago

6 0

Answer:

Air has greater density.

The object will float.

Explanation:

Density of a body is defined as the mass of a body per unit volume of it.

Here, air has a density of 1.27 kg per ml and Helium balloon has a density of 0.33 kg per ml. This means that if volume is 1 ml, then the mass of air in that volume will be 1.27 kg and mass of Helium balloon in that same volume will be 0.33 kg.

Therefore, the mass of Helium balloon is lighter than that of air in a given volume. Hence, air is heavier and has a greater density.

A lighter density object always floats over a greater density body.

Hence, Helium balloon will float over air as Helium balloon is less dense than that of air.

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A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie

-BARSIC- [3]

a) $3.27\cdot 10^{-3} J$

b) $11.60\cdot 10^{-3} J$

c) $8.33\cdot 10^{-3} J$

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

$C=13.5 \mu F = 13.5\cdot 10^{-6}F$ is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

$U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J$

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

$C'=kC$

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

$U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2$

where:

$C=13.5\cdot 10^{-6}F$ is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

$U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J$

The initial energy stored in the capacitor, before the dielectric is inserted, is

$U=3.27\cdot 10^{-3} J$

The final energy stored in the capacitor, after the dielectric is inserted, is

$U'=11.60\cdot 10^{-3} J$

Therefore, the change in energy of the capacitor during the insertion is:

$\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J$

So, the energy of the capacitor has increased by $8.33\cdot 10^{-3} J$

8 0

3 years ago

give three examples from the device on the apply page where potential energy was converted to kinetic energy

Elodia [21]

a water turbine drops water onto a blade which gains ke at water's gpe expense

4 0

3 years ago

PLEASE HELP I NEED TO TURN IT IN IN AN HOUR ITLL GIVE YOU POINTS PLS PLEASE

miv72 [106K]

Answer:

17.95025
172.3995

Explanation:

$\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{2.995}{0.16685}\:\\\mathrm{is}\:17.95025$

$\mathrm{Multiply\:without\:the\:decimal\:points,\:then\:put\:the\:decimal\:point\:in\:the\:answer}\\\\910\times\:18945=17239950\\\\910\mathrm{\:has\:}0\mathrm{\:decimal\:places}\\0.18945\mathrm{\:has\:}5\mathrm{\:decimal\:places}\\\\\mathrm{Therefore,\:the\:answer\:has\:}5\mathrm{\:decimal\:places}\\\\=172.39950\\\\Refine\\=172.3995$

5 0

3 years ago

A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is

Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

P₁ V₁ = P₂ V₂

P_atm V₁ = (P_atm + ρ g H) V₂

V₂ = V₁ P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

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3 years ago

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mars1129 [50]

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3 years ago