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creativ13 [48]
2 years ago
8

batman recharges his bat taser with 2500 joules of energy in 15 minutes. how much power does it draw from the mains

Physics
1 answer:
Yuliya22 [10]2 years ago
6 0
The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):
P(W) = E(J) / t(s)

2.78 Watts
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A 2-Newton upward net force is being applied to a 10-kg object. What is the magnitude of the upward
Colt1911 [192]
Acceleration in m/s^2 = 2/10 = 0.2 m/s^2
7 0
3 years ago
A solid object has a mass of 104 kg and a volume of 1,278 m3. What is its density?
MrMuchimi
The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.
4 0
3 years ago
Does aluminum foil refract light? Why?
nikitadnepr [17]

yes it does it because the material cause it to refract and the stronger it is the more it will refract

4 0
3 years ago
How is heat converted into work in a steam engine?
Dmitry_Shevchenko [17]
In an extremely simplified explanation,
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8 0
3 years ago
Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation
Artemon [7]

Answer:

L =   μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

        E_{L} = - L dI / dt

 

Let's use Faraday's law

       E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

        - d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

        n d Ф_B = L dI

we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

       L =   μ₀ n r / 2I

4 0
3 years ago
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