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4 years ago

Pressure in a hydraulic system is achieved by applying the force of gravity on a gas.

Physics

2 answers:

lana66690 [7]4 years ago

7 0

B) False.

The force of gravity is not used to create the hydraulic force.

-Dominant- [34]4 years ago

5 0

Your answer would be B. False. Due to in a hydraulic system, you can have pressure put on gas without gravity as well. Think of it like a shock absorber on a car, when you hit a bump in the road, the shock compresses, not due to gravity, but due to the weight of the car pushing down, while the wheel is pushing up. If anything, it would be closer to maybe, the law of inertia.
Hope this helps!

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Explain how a bathroom scale is like a biofeedback machine.

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Answer:

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3 years ago

Nuclear Energy definition

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Answer:

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3 years ago

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 128 ∘C. The gas expands and, in the process, a

o-na [289]

Answer:

The final temperature of the gas is <em>114.53°C</em>.

Explanation:

Firstly, we calculate the change in internal energy, ΔU from the first law of thermodynamics:

ΔU=Q - W

ΔU = 1180 J - 2020 J = -840 J

Secondly, from the ideal gas law, we calculate the final temperature of the gas, using the change in internal energy:

$ΔU=\frac{3}{2} nRΔT$

$ΔU=\frac{3}{2} nR(T_{2} -T_{1} )$

Then we make the final temperature, T₂, subject of the formula:

$T_{2} =\frac{2ΔU}{3nR} +T_{1}$

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Therefore the final temperature of the gas, T₂, is 114.53°C.

7 0

4 years ago

Ms. Sparkle bought 12 cans of diet soda. Each can

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4.2 liters..... there are 1,000 mL in a liter and there is a total of 4200 mL in this case which is divided by 1000 which gives you 4.2 liters.

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3 years ago

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

6 0

3 years ago