Ask question

Ask question

All categories

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?

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

You might be interested in

Interpret the following graphs by answering the questions below.

Furkat [3]

Answer:

Nodes = 5 and antinodes = 4

Explanation:

Nodes are the points of zero amplitude and appear to be fixed. On the other hand, antinodes are points on a stationary wave that oscillates with maximum amplitude.

In this given standing wave, there are 5 points where the amplitude is 0. So, there are 5 nodes. Also, there are 4 points where the amplitude is maximum.

So, there are 5 nodes and 4 antinodes.

6 0

2 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

<u></u>

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

What sort of fitness component would dance be<br> categorized under

zalisa [80]

Answer:

aerobic fitness

Explanation:

i just looked it up lol

6 0

3 years ago

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0

3 years ago

What was the average velocity for the entire trip?

just olya [345]

Answer:

<u><em>3.721 m/s</em></u>

This is the explanation of the ans

6 0

2 years ago