The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
The given parameters;
- <em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
- <em>Distance between each floor, d = 10 ft</em>
The vertical pressure of the water is calculated as follows;
The vertical height of the first floor from the 13th floor = 130 ft
The vertical height of the 13 ft floor = 10 ft
Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
Learn more about vertical height and pressure here: brainly.com/question/15691554
Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N
Answer:
The time for final 15 cm of the jump equals 0.1423 seconds.
Explanation:
The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as
where
'v' is the final velocity of the player
'u' is the initial velocity of the player
'a' is acceleration due to gravity
's' is the height the player jumps
Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get
Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as
Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as
where symbols have the usual meaning
Applying the given values we get
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 ?
Answer:
The answer to your question will be d) greater the pressure
Explanation:
