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1 year ago

What potential difference is required across an 32 -Ω resistor to cause 33.72 A to flow through it?

Physics

1 answer:

Inessa05 [86]1 year ago

6 0

$\text{Voltage,}~ V = IR =33.72\times 32 = 1079.04~ \text{volt}$

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6. several light bulbs are connected in series across a 115 v source

Natalka [10]

Explanation:

a) I = V / R

1.70 = 115 / R

R = 115 / 1.70

R = 67.647

R = 67.65 ohms

Therefore, equivalent resistance is 67.65 ohms

b) Equivalent resistance of circuit from above sum is 67.65 ohms

Given resistance of each bulb is 1.50 ohms

Number bulbs = Equivalent resistance / Resistance of each bulb

= 67.65 / 1.50

= 45

8 0

1 year ago

A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1

stellarik [79]

Answer:

563712.04903 Pa

Explanation:

m = Mass of material = 3.3 kg

r = Radius of sphere = 1.25 m

v = Volume of balloon = $\frac{4}{3}\pi r^3$

M = Molar mass of helium = $4.0026\times 10^{-3}\ kg/mol$

$\rho$ = Density of surrounding air = $1.19\ kg/m^3$

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

$mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg$

Mass of helium is 6.4356 kg

Moles of helium

$n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489$

Ideal gas law

$P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa$

The absolute pressure of the Helium gas is 563712.04903 Pa

3 0

2 years ago

Unpolarized light with an average intensity of 845 W/m2 enters a polarizer with a vertical transmission axis. The transmitted li

RideAnS [48]

The concept to develop this problem is the Law of Malus. Which describes what happens with the light intensity once it passes through a polarized material.

Mathematically this can be expressed as

$I = I_0 cos^2\theta$

Where

I = New intensity after pass through the Polarizer

$I_0$ = Original intensity

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

When the light passes perpendicularly through the first polarizer, the light intensity is reduced by half which will cause the intensity to be $225W / m ^ 2$ at the output of the new polarizer, mathematically: