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postnew [5]
3 years ago
12

I need to lift a 2000kg car, 1.798m and the joules required is 35240.8. Converted to watt (W = 35240.8/5 (s)) I got 7048.16 W. I

also have a 14.4 volt battery. Does the battery hold enough energy to lift the car if powering a lifting machine?
Physics
1 answer:
marusya05 [52]3 years ago
5 0
This is a very interesting problem ... mainly because it's different from
the usual questions in the Physics neighborhood.

I can discuss it with you, but maybe not quite give you a final answer
with the information you've given in the question.

I agree with all of your calculations so far ... the total energy required,
and the power implied if the lift has to happen in 5 seconds.

First of all, let's talk about power.  I'm assuming that your battery is
a "car" battery, and I'm guessing you measured the battery voltage
while the car was running.  Turn off the car, and you're likely to read
something more like 13 to 13.8 volts.
But that's not important right now.  What I'm looking for is the CURRENT
that your application would require, and then to look around and see whether
a car battery would be capable of delivering it.

   Power = (volts) x (current)

   7,050 W  =  (14 volts) x (current)

   Current = (7,050 watts / 14 volts) =  503 Amperes. 

That kind of current knocks the wind out of me.  I've never seen
that kind of number outside of a power distribution yard.
BUT ... I also know that the current demand from a car battery during
starting is enormous, so I'd better look around online and try to find out
what a car battery is actually capable of.

I picked a manufacturer's name that I'd heard of, then picked their
recommended battery for a monster 2003-model car, and looked at
the specs for the battery.

The spec I looked at was the 'CCA' ... cold cranking Amps.
That's the current the battery is guaranteed to deliver for 30 seconds,
at a temperature of 0°F, without dropping below 12 volts.

This battery that I saw is rated  803 Amps  CCA !

OK.  Let's back up a little bit.  I'm pretty sure the battery you have
is a nominal "12-volt" battery.  Let's say you use to start lifting the lift. 
As the lift lifts, the battery voltage sags.  What is the required current
if the battery immediately droops to 12V and stays there, while delivering
7,050 watts continuously ?

          Power = (volts) x (current)

          7,050 W = (12 V) x (current)

            Current = (7,050 W / 12 V)  =  588 Amps . 

Amazingly, we may be in the ball park.
If the battery you have is rated by the manufacturer for 600 Amps
CCA (0°F) or CA (32°F), then the battery can deliver the current
you need.
BUT ... you can't conduct that kind of current through ear-bud wire,
or house wiring wire.  I'm not even so sure of jumper-cables. 
You need thick, no-nonsense cable, AND connections with a lot of
area ... No alligator clips.  Shiny nuts and bolts with no crud on them.

Now ... I still want to check the matter of the total energy.
I'm sure you're OK, because the CCA and CA specifications talk about
30 seconds of cranking, and you're only talking about 5 seconds of lifting.
But I still want to see the total energy requirement compared to the typical
battery specification ... 'AH' ... ampere-hours.

You're talking about 35,000 joules

                          = 35,000 watt-seconds

                         =  35,000 volt-amp-seconds.

               (35,000 volt-amp-sec) x (1 hour/3600 sec) / (12 volt)               

           =  (35,000 x 1) / (3600 x 12)  volt-amp-sec-hour / sec-volt

           =    0.81 Amp-Hour  .

That's an absurdly small depletion from your car battery.
But just because it's only  810 mAh, don't get the idea that you can
do it with a few rechargeable AA batteries out of your camera.
You still need those 600 cranking amps.  That would be a dead short
for a stack of camera batteries, and they would shrivel up and die.

Have I helped you at all ?
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A 57 kg wagon is pulled with a constant net force of 38 N. Calculate the acceleration of the wagon. F = ma
Dahasolnce [82]

Answer:

<h2>0.67 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{38}{57} =  \frac{2}{3}   \\  = 0.666666...

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

4 0
2 years ago
A ball is traveling 24° above the horizontal at a speed of 12 m/s. What is the vertical component of its speed?
victus00 [196]

Answer:

4.88 m/s

Explanation:

Vertical component would be  12 * sin 24  =  4.88 m/s

Horizontal is   12 * cos 24

3 0
2 years ago
On my science test, there is a bonus question that I want to get right. Why would it be a bad idea to skydive on the moon? Hint:
eduard

Answer:

Because there is no air resistance

Explanation:

When an object falls on Earth, there are essentially two forces acting on it:

- The force of gravity, downward, equal to the weight of the object:

W=mg

where m is the mass and g the acceleration due to gravity

- The air resistance, F, which acts upward, and whose magnitude is generally proportional to v, the speed of the object

When the object starts its fall, its initial speed is zero: v = 0, so the air resistance is also zero: F=0, and the object accelerates downward due to gravity.

However, as it accelerates downward, its speed increases, and so does the air resistance F. However, F is upward, opposite to the direction of motion, therefore it reduces the net acceleration of the object; at a certain point, the magnitude of the air resistance will become equal to the weight, so that

mg = F

and at that point, the net acceleration of the object will become zero: this means that the object will continue its fall at a constant velocity, called terminal velocity.

On the Moon instead, there is no air resistance: this means that for an object falling down, the speed keeps increasing due to the effect of gravity, and it will never reach a terminal value: therefore, the final velocity at the impact will be much higher than on the Earth, if we assume the two objects have been dropped from a very high altitude from the surface.

7 0
3 years ago
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

\frac{y-x}{y}=\frac{6}{15}

15 y - 15 x = 6 y

9 y = 15 x

y=\frac{5}{3}x

Differentiate with respect to time.

\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}

As given, dx/dt = 4 ft/s

\frac{dy}{dt}=\frac{5}{3}\times 4

\frac{dy}{dt}=\frac{20}{3} ft/s

6 0
3 years ago
A ball dropped from a window strikes the ground 2.76 seconds later. How high is the window above the ground
sweet [91]

Answer:

37.33m

Explanation:

Using the equation of motion

S = ut + 1/2gt^2

Time t = 2.76secs

g = 9.8m/s^2

S = 0 + 1/2(9.8)(2.76)^2

S = 4.9*7.6176

S = 37.33

Hence the window is 37.33m above the ground

7 0
3 years ago
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