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tatuchka [14]
2 years ago
11

Example: How much pressure is created when you apply a 5.00 N force on a syringe plunger that has a diameter of 2.00 cm

Physics
1 answer:
nekit [7.7K]2 years ago
4 0

The pressure created when you apply a 5.00 N force on a syringe plunger that has a diameter of 2.00 cm is 7961.78 \frac{N}{m^2}.

Solution:

Pressure = \frac{force}{area}

P = \frac{5}{\pi \times\ 0.02^2}

P = 7961.78 \frac{N}{m^2}

<h3>What is pressure?</h3>
  • The force perpendicular to an object's surface that is delivered per unit area over that force is known as pressure (symbol: p or P). Gauge pressure, often known as gage pressure, is the pressure in relation to the surrounding air pressure.
  • To express pressure, a variety of units are employed. The SI unit of pressure, the pascal (Pa), for instance, is equal to one newton per square meter (N/m2).
  • The pound-force per square inch (psi), on the other hand, is the traditional unit of pressure in the imperial and U.S. customary systems. Standard atmospheric pressure (atm) is another way to express pressure; it is the same as atmospheric pressure.

To learn more about pressure with the given link

brainly.com/question/356585

#SPJ4

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The work done to lift the counterweight equals the potential energy acquired

Explanation:

since this is vertically applied force on the counterweight, and the distance the force is displacing the counterweight is in the same direction as the applied force, it equals the gained potential energy

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The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge
Eva8 [605]

Answer:

a) v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)           (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}

We can write the mass of a proton in MeV/c².

m_{p}=938.28 MeV/c^{2}

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.04

v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) Linac (400 MeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.71

v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) Booster (8 GeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.994

v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) Main ring or injector (150 Gev)

\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.999

v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) Tevatron (1 TeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.9999

v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Have a nice day!

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Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

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Ty's blood is slightly alkaline

Explanation:

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