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elena-s [515]
3 years ago
9

Which animal have no teeth​

Physics
2 answers:
Kamila [148]3 years ago
6 0
Anteaters they do not have teeth
dmitriy555 [2]3 years ago
5 0

Answer:

Ant eaters

Explanation:

Pls give me brainleist

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A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J
cestrela7 [59]

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = \frac{energy}{time}

The energy needed to light one bulb=power*time

Given time = 1minute = 60 seconds

Energy = 32W*60sec=1920J

Therefore energy needed to light one bulb is 1920J

The energy needed to light 24 bulbs = 1920*24 =46080J=46.08KJ

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3 years ago
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Which of the following energies defines energy stored inside a spring?
Leya [2.2K]

Answer:

Elastic potential energy

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2 years ago
At which points would the river be traveling the slowest?
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Answer:

This would be traveling at the lower reaches.

Explanation:

A river would be traveling the fastest at the upper reaches and it becomes slower at the middle reaches and the slowest at the lower reaches. A place where water flows fast in a river is where the width is narrow and the bottom is steep. (This is just examples incase you would like to keep notes).

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3 years ago
If you were told an atom was an ion, you would know the atom must have a ______?
andreyandreev [35.5K]

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charge

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3 years ago
On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi
Lostsunrise [7]

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

13=152z+7.6\times\frac{13-180z}{4.5}

which simplifying gives us, z=0.0589\Omega/m\approx0.06\Omega/m (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

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2 years ago
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