Answer:
B
Explanation:
Water level remains unchanged
The first thing to realize is that the buoyancy force is the same as, or equal to the weight of the wood, this same force is also the same as or equal to the weight of the water displaced by the wood. In the two cases, the weight of the wood will be unaffected nonetheless, and thus the water level will remain the same.
Therefore, the answer is B, the water level remains unchanged.
Answer:
We should not consider the needs of humans first because Conservation of nature is more important.
Explanation:
Human beings are important but the priority must be given to nature. Human beings have uncountable needs and if we focus on satisfying all such needs then it will affect our nature. Because finally our needs are fulfilled by using the resources from nature.
So rather than considering the needs of humans first, we must try to preserve our nature as much as possible. It is not done then it will lead to disaster.
Answer:
The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.
Explanation:
Data
F = 7 N
F = ? if the masses is quartered
Formula
Process
Normal conditions F = Km₁m₂/r² = 7
When masses quartered F = K(m₁/4)(m₂/4)/r² = ?
F = K(m₁m₂/16)/r²
F = K(m₁m₂/16r² = 7/16 = 0.4375 N
The power of man performing 500 J of work in 8 seconds is 62.5 J/s.
Power can be defined as the pace at which work is completed in a given amount of time.
Horsepower is sometimes used to describe the power of motor vehicles and other machinery.
The pace at which work is done on an item is defined as its power. Power is a temporal quantity.
Which is connected to how quickly a project is completed.
The power formula is shown below.
Power = Energy / Time
Power = E / T
Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.
Here we have given Energy as 500 J and Time as 8 second.
Power = Energy / Time
Power = 500 / 8 Joule / sec
Power = 250 / 4 Joule / sec
Power = 125 / 2 Joule / sec
Power = 62.5 Joule / sec or 62.5 watt
Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.
So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.
Learn more about Power here:
brainly.com/question/1634438
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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 ?
