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2 years ago

What is the length of a pendulum that has a period of 0.500s?

Physics

1 answer:

anyanavicka [17]2 years ago

7 0

That is the answer to your problem

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GarryVolchara [31]

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2 years ago

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What work do you think is done when you carry a 20 N weight backpack for a 1000 m walk? will the work be positive, negative, or

dezoksy [38]

The work done is positive and is equal to 20000 J

Work done is defined as the product of force and the distance moved by the force.

Mathematically:

Work done = force * distance

The work done by the force = 20 * 1000 = 20000J

The work done is positive and is equal to 20000 J

Learn more about work done at: brainly.com/question/25923373

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1 year ago

A soccer ball is kicked 8 m north. Then a teammate kicks it 6 m east into the goal. What is the soccer ball’s displacement from

dimulka [17.4K]

Answer:

the displacement is zero and the distance is 100 meters

Explanation:

7 0

2 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

7 0

2 years ago

Please help me with this question

drek231 [11]

Hi, here is the first part of the answer..do you know what type of graph it's wanting?? (line graph, bar graph, ect.)

5 0

2 years ago