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Black_prince [1.1K]
3 years ago
11

A type of wave that carries energy from one place to another, even through

Physics
1 answer:
Igoryamba3 years ago
8 0

Answer:

light waves :)

Explanation:

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A 60.0-kg person is in an elevator with a mass of 2000 kg. The elevator moves vertically up through a distance of 24.0 m with a
kati45 [8]

The mass of the (elevator + person) is (2,000 kg + 60 kg) = 2,060 kg .

The weight is (mass x gravity) = (2,060 x 10) = 20,600 newtons

Work = (force x distance) = (20,600 newtons x 24m) = <em>494,400 joules</em>

The speed, acceleration, and how much time it takes don't make any
difference, unless you want the average power during the lift.

         Power = (work)/(time) =
           494,000 J / (24/4 m) =
           494,000 J / 6 sec = <em><u>82,400 watts</u></em>      wow !
6 0
3 years ago
Earths gravitational pull just got 3 times stronger what happens to your weight?
djverab [1.8K]
You get 3 times heavier
6 0
3 years ago
Read 2 more answers
Without using a micrometer screw gauge, how do I find the average diameter of a long piece of thin wire using a metre rule and a
Mice21 [21]

Answer:

Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.

Explanation:

Since the diameter of one long piece of thin wire is too thin to be measured by a metre ruler, you can wind it multiple times and push it side by side to get a length you can measure.

For example, if you wound it around 20 times and the total length of 20 diameters of the wire side-by-side is 2.0 cm, one winding, which is the diameter would be 2.0cm ÷ 20 = 0.10cm or 1mm.

5 0
2 years ago
V=I/R correctly expresses the relationship between voltage, current, and resistance.
docker41 [41]
The correct answer is true
5 0
2 years ago
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How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800
ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2

The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0
3 years ago
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