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Rama09 [41]
3 years ago
6

“ to run up the stairs the student must work against the force of ....”

Physics
1 answer:
Korolek [52]3 years ago
7 0
The answer here would be the force of gravity Hope I could help! :)
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What are the properties of water from a glacier and a geyser that are the same
Sav [38]
The only common property is that both of those things are comprised of H2O.
7 0
4 years ago
Water drops from the nozzle of a shower onto the floor 81 inches below. The drops fall at regular intervals of time, the first d
Alexxx [7]

Answer:

0.91437 m

0.22859 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s=81\ inches=81\times 0.0254=2.0574\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.0574=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.0574\times 2}{9.81}}\\\Rightarrow t=0.64764\ s

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.64764}{3}\\\Rightarrow t'=0.21588\ s

For second drop time is given by

t''=2t'\\\Rightarrow t''=2\times 0.21588\\\Rightarrow t''=0.43176\ s

Distance from second drop

s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.43176^2\\\Rightarrow s=0.91437\ m

Distance from second drop is 0.91437 m

Distance from third drop

s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.21588^2\\\Rightarrow s=0.22859\ m

Distance from third drop is 0.22859 m

6 0
3 years ago
Properties of helium
postnew [5]

Answer:

Helium has many unique properties: low boiling point, low density, low solubility, high thermal conductivity and inertness, so it is use for any application which can explioit these properties. Helium was the first gas used for filling balloons and dirigibles

7 0
3 years ago
What changes occur in the molecular magnet when magnet is demagnetized​
Olin [163]

Answer:

A magnet is demagnetized. There is a decrease in chemical potential energy. There is an increase in thermal energy. And there is an increase in energy in the form of demagnetization.

8 0
3 years ago
Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F
muminat

Answer:

v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)

Explanation:

Given that the force of the particle is,

F_{x}=F_{0}e^{-kt}

Now it can be further written as

m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C

Now the initial conditions are v=1 at t=0.

So,

1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}

Now the velocity will become.

v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

And,

\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\

And, another initial condition is x=0 at t=0

0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}

Now,

x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)

5 0
3 years ago
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