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Pepsi [2]
3 years ago
5

A battery charger is connected to a dead battery and delivers a current of 8.9 A for 4.7 hours, keeping the voltage across the b

attery terminals at 12 V in the process. How much energy is delivered to the battery?
Physics
2 answers:
andrew11 [14]3 years ago
5 0

Answer:

1807056 J

Explanation:

Energy: This can be defined as the ability or the capacity to do work. The S.I unit of energy is Joules(J).

The expression for electric energy is given as,

Using,

E = VIt...................... Equation 1

Where E = Electric Energy, V = Voltage, I = current, t = time.

Given: V = 12 V, I = 8.9 A, t = 4.7 hours = 4.7×3600 = 16920 s

Substitute into equation 1

E = 12(8.9)(16920)

E = 1807056 J.

Hence the amount of energy delivered to the battery = 1807056 J

adell [148]3 years ago
3 0

Answer:

1807.56 kJ

Explanation:

Parameters given:

Current, I = 8.9A

Time, t = 4.7hrs = 4.7 * 3600 = 16920 secs

Voltage, V = 12V

Electrical energy is given as:

E = I*V*t

Where I = Current

V = Voltage/Potential differenxe

t = time in seconds.

E = 8.9 * 12 * 16920

E = 1807056 J = 1807.056 kJ

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A stunt woman attempts to swing from the roof of a m 24 high building to the bottom of an identical building using a m 24 rope,
MakcuM [25]

Answer:

Height from ground is 8 m where string will break

Explanation:

Let the string makes some angle with the vertical after some instant of time

So here we have

T - mg cos\theta = \frac{mv^2}{R}

T = mg cos\theta + \frac{mv^2}{R}

now by energy conservation we have

\frac{1}{2}mv^2 = mg(R cos\theta)

mv^2 = 2mgR cos\theta

T = mg cos\theta + 2mgcos\theta

T = 3mg cos\theta

For string break down we have

T = 2mg = 3mgcos\theta

cos\theta = \frac{2}{3}

Now height from the ground is given as

h = R - Rcos\theta

h = 24 - 24(\frac{2}{3})

h = 8 cm

7 0
3 years ago
A father pushes his child on a swing. He pushes with a force of 20 N or a distance of 1 m, with the force always maintained para
Firlakuza [10]

Answer:

20 Joules

Explanation:

Work is done whenever a force moves a body through a certain distance in the direction of the force. So, work done is the product of force and distance moved.

Therefore, we have;

Work done = Force x distance

i.e   Wd = Fs

Given that: F = 20 N and s = 1 m, then;

Wd = 20 N x 1 m

     = 20 Nm

The work done by the father is 20 Joules(Nm).

7 0
3 years ago
a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the
erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em>  has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

W=Fx

This is equal to the change in the bullet's kinetic energy.

W=Fx=\frac{1}{2} m(u^2-v^2)......(1)

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s

Thus the speed of the bullet is 71 m/s


3 0
3 years ago
Jane believes that plate tectonics is a theory, and Mario believes it is a law. Which of the following best supports Jane’s argu
Gekata [30.6K]

Answer:

its d

Explanation:

i did this and got it right

8 0
3 years ago
Read 2 more answers
1. An electric iron has a rating of 750W, 220V. Calculate:
horrorfan [7]
The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a). 6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span> For 7, realize that because they are in parallel, their voltages are the same. We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span> For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>
6 0
3 years ago
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