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Assoli18 [71]
3 years ago
15

Where does the energy to turn a flashlight on come from

Physics
2 answers:
Ne4ueva [31]3 years ago
8 0

Answer:

Electrical energy is the energy of electricity—electrical charges moving through a conductor. Electricity can be transported through wires to places where it is needed and then converted into other forms. In a flashlight, the electrical energy becomes light energy and thermal energy in the bulb.

Explanation:

Anastasy [175]3 years ago
5 0

Answer:

the battery

Explanation:

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What is the kinetic energy of a 50-kg child running to catch the school bus at
Vinvika [58]

Answer:

Option C

100 J

Explanation:

Kinetic energy, KE is given by

KE=0.5mv^{2} where m is the mass and v is the velocity

Substituting 50 Kg for mass, m and 2 m/s for velocity v then we obtain

KE=0.5*50*2^{2}=100 J

Therefore, the child's kinetic energy is equivalent to 100 J

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3 years ago
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False

mid-ocean ridge

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3 years ago
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What the planet after Venus
Aleonysh [2.5K]

Mercury and Venus are therefore closer to each other most of the time. But Earth is the planet closest to Venus. And that's why from here on Earth, Venus looks so big and luminous. Venus is the brightest thing in the night sky after the sun and the moon.

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3 years ago
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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
The plate near the Artic Ridge is among those with the slower growth rates, moving slightly less than 2.5 cm/year. How much will
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Greater than 25 . 2.5 x 10 = 30
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