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Bogdan [553]
2 years ago
5

pressure A vertical piston cylinder device contains a gas at an unknown pressure. If the outside pressure is 100 kPa, determine

(a) the pressure of the gas if the piston has an area of 0.2 m2 and a mass of 20kg. Assume g = 9.81 m/s2. (b) What-if Scenario: What would the pressure be if the orientation of the device were changed and it were now upside down

Physics
1 answer:
ycow [4]2 years ago
5 0

Answer:

Explanation:

When Cylinder is vertical Let gas pressure inside is P_1 and atmospheric Pressure be

P_0=100 KPa

Area of Piston A=0.2 m^2

mass of Piston m=20 kg

from FBD

mg+P_0A=P_1A

P_1=P_0+\frac{mg}{A}

P_1=100+\frac{20\times 9.8}{0.2}

P_1=100+0.98

P_1=100.98 KPa

(b)If cylinder is upside down

From FBD

mg+P_2A=P_0A

P_2=-\frac{mg}{A}+P_0

P_2=100-\frac{20\times 9.8}{0.2}

P_2=100-0.98=99.02 KPa

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According to newton's third law of motion, when a hammer strikes and exerts force to push it into a piece of wood, the nail
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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e
antoniya [11.8K]

Answer:

r=5.278\times 10^{-4}\ m

Explanation:

Given that:

  • magnetic field intensity, B=0.07\ T
  • kinetic energy of electron, KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J
  • we have mass of electron, m=9.1\times 10^{-31}\ kg

<em>Now, form the mathematical expression of Kinetic Energy:</em>

KE= \frac{1}{2} m.v^2

1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2

v^2=4.2198\times 10^{11}

v=6.496\times 10^6\ m.s^{-1}

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

r=\frac{m.v}{q.B}

r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}

r=5.278\times 10^{-4}\ m

6 0
3 years ago
A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
2 years ago
how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0
2 years ago
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