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3 years ago

11

A 7.4 A current is set up in a circuit for 7.8 min by a rechargeable battery with a 3.0 V emf. By how much is the chemical energ

y of the battery reduced?

1 answer:

bearhunter [10]3 years ago

8 0

Answer:

10389.6 J

Explanation:

Power is the rate of doing work with respect to time, its S.I unit is in watts but it can also be expressed in J/s. Power is calculated using the formula:

$Power=\frac{energy}{time}$

Power is also the rate at which energy is used per second.

Energy is the capacity to do work and it is measured in joules (J).

Power = current × voltage = 7.4 A × 3 V = 22.2 W

$Energy=power*time\\\\time=7.8\ min=7.8*60\ s=468\ s\\\\Energy=22.2*468\\\\Energy=10389.6\ J$

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Setler79 [48]

Answer:

L = 7 [cm]

Explanation:

To solve this problem we must analyze each of the distances mentioned and take into account the number of covers and thicknesses of these.

The worm crosses through the sheets of the first book, this distance can be determined by the following length analysis.

4 = P + 2*C

Where:

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P = 4 - 2*(0.5)

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L = P + 2C + P "the pages of the first book + 2 covers + the pages of the second book"

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3 years ago

Which of the following compounds does not contain molecules? A. H2 B. NaCl C. CO2 D. H2O

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B. NaCl

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4 years ago

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Kelly sits on a rock. Her weight is an action force. Describe the reaction force.

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An equal and opposite force acting upwards

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A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

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Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

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ehidna [41]

Answer:

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Explanation:

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