Force exerted by a person or object is called

ikadub [295]

The first law of thermodynamics states the conservation of energy and heat where the total energy in an isolated system may be transformed into another, but never created or destroyed. If 286 J of energy was released to the room, then also 286 J of energy was also removed from food in that refrigerator assuming it is an isolated system. :)

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3 0

3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

3 years ago

HELP ASAP‼️<br><br> Can someone please answer this question I’m confusion❔

klemol [59]

the enclosed may help. earth rotes and you and i are fixed to it. air flows it's own path ...

3 0

3 years ago

45 grams in kilo grams

Hunter-Best [27]

Answer:

0.045 kg

Explanation:

1000 grams = 1kg

7 0

3 years ago

Read 2 more answers

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

marusya05 [52]

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 ?

5 0

3 years ago