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2 years ago

A light bulb does 100 joules of work in 2.5 seconds how much power does it have

Physics

2 answers:

Alborosie2 years ago

8 0

Answer: 40Watt

Explanation: Power is the amount of work done in one seconds. Also know as 1Watt.

Therefore, if the light bulb consume an energy of 100joule in 2.5 secs. It means that the energy which is equal to work done in 1secs would be,

100/2.5=40 joules in one second which is the Power. Which can also be written in the standard unit of power {Watt} as 40Watt.

Vladimir79 [104]2 years ago

6 0

4 just divide 100 to 25 which is 4

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A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to m

Umnica [9.8K]

Answer:

Explanation:

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2 years ago

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = $\frac{4}{3} \pi (\frac{D}{2})^3$

Therefore $\pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3$

r of the disc = $1.15(\frac{ D}{2} )$

Using conservation of angular momentum;

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2 years ago

Calculate the applied force of the washers on the car. First, convert the mass you recorded for one, two, three, and four washer

Andrej [43]

The mass of one washer is 0.0049 kg.

The mass of two washers is 0.0098 kg.

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2 years ago

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What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i

Katarina [22]

R = 0.407Ω.

The resistance R of a particular conductor is related to the resistivity ρ of the material by the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4. Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

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2 years ago

Which of the following statements cannot be supported by Kepler's laws of planetary motion?

gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- A planet's distance from the sun will not be the same in six months. --> using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

- A planet's speed as it moves around the sun will not be the same in six months. --> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. --> using the 3rd law. In fact, the ratio $\frac{a^3}{T^2}$ (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Is not supported by any Kepler's law.

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2 years ago