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vivado [14]
3 years ago
5

How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?

Physics
1 answer:
salantis [7]3 years ago
3 0

Answer:

Power = 20 Watts

Explanation:

Given the following data;

Voltage = 100 V

Resistance = 500 Ohms

To find the power that is required to light a lightbulb;

Mathematically, power can be calculated using the formula;

Power = \frac {Voltage^{2}}{resistance}

Substituting into the formula, we have;

Power = \frac {100^{2}}{500}

Power = \frac {10000}{500}

Power = 20 Watts

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An electrician wraps rubber electrical tape around a copper wire. Since rubber is an insulator, which statement best explains th
yuradex [85]

A) Rubber stops charges from flowing. This protects people by stopping electricity from flowing.

Explanation:

The statement that best describes the point of wrapping rubber around the copper wire is that the rubber stops charges from flowing. This prevents people from getting electrical shocks by stopping the flow  of electricity.

  • A rubber is an insulator.
  • Insulators are substances that prevents the flow of electricity.
  • The lack free mobile electrons or ions that makes them conductors.
  • When they are wrapped round a conductor such as copper wire, they will halt the flow of charges.
  • Copper is a conductor of both heat and electricity. It has free mobile electrons.

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4 0
4 years ago
Anyone.... help with this two questions...​
PSYCHO15rus [73]

Answer:

what is the question

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Explanation:

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7 0
3 years ago
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maria [59]

Answer:

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4 0
3 years ago
A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an
Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

R is resistance

The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

So

ΔV=IR

=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V

ΔV=0.484mV  

3 0
3 years ago
A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the stri
frutty [35]

Answer:

-2.26×10^-4 radians

Explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians

8 0
4 years ago
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