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frutty [35]
3 years ago
15

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours (A⋅h). A 50 A⋅h

battery can supply a current of 50 A for 1.0 h, or 25 A for 2.0 h, and so on.
What total energy can be supplied by a 13 V, 60A?h battery if its internal resistance is negligible?
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

2.80 MJ  

Explanation:

(a) We want to calculate the energy U of the battery, where its voltage is E = 13.0V and the supplied current is I = 60 A. We can neglect the internal resistance, so the terminal voltage equals the emf of the battery V = 13.0V. The quantity of delivered energy is given by the rate at which energy is delivered to it in a certain time t. We could obtain the rate at which energy is transferred by using equation , where the rate represents the power P = IV. Therefore, the energy produced is given by  

U = P*t                                                (P = IV)            

U = I*V*t                                                                     (1)  

Now we can plug our values for I, V and t into equation (1) to get the energy produced in time t = 1 h = 3600 s  

U = I*V*t = (60 A)(13 V)(3600s) =  2.80 MJ  

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After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

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5 0
3 years ago
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

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We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
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neonofarm [45]

Answer:

true

Explanation:

as long as it is the right conductive material

5 0
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