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Digiron [165]
3 years ago
13

What happens to the strength of an electromagnet if the number of loops of wire is increased?

Physics
1 answer:
luda_lava [24]3 years ago
8 0

Answer:

If you coil the wire around and around, it will make the magnetic force stronger, but it will still be pretty weak. ... The strength of an electromagnet can be increased by increasing the number of loops of wire around the iron core and by increasing the current or voltage.

Explanation:

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Is the substance an atom or a molecule?
ANTONII [103]

Answer:

A pair of oxygen atoms is a molecule of oxygen. A molecule is the smallest particle of a substance that exists independently. Molecules of most elements are made up of only one of atom of that element. Oxygen, along with nitrogen, hydrogen, and chlorine are made up of two atoms.

Explanation:

6 0
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A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from
ladessa [460]

Answer:

P_1 = 1166.7 Watt

P_2 = 2000 Watt

Explanation:

Average power for the human sprinter is given as

Power = \frac{\Delta E}{\Delta t}

so we have

P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}

P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}

P_1 = 1166.7 Watt

Average power for greyhound is given as

P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}

P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}

P_2 = 2000 Watt

3 0
3 years ago
Deaths due to lifestyle diseases have declined since the early 1900s.<br><br> T<br> F
zavuch27 [327]

Answer:

<h2>False</h2>

Explanation:

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7 0
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O be useful in science, a hypothesis must be *
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Answer:

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3 0
3 years ago
Read 2 more answers
A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?
garri49 [273]
<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
6 0
3 years ago
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