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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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A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km
ruslelena [56]

Answer:

x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

Angle which makes from ground = 58.9°

(a) X component of distance  is given by r_x=rcos\Theta =118\times cos58.9=118\times 0.5165=118=60.85m

(b) Y component of distance is given by r_Y=rcos\Theta =118\times sin58.9=118\times 0.8562=101.0316m

These are the x and y component of position vector

6 0
3 years ago
A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0
2 years ago
A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended
Alik [6]
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
k= \frac{F}{\Delta x}
where F is the force applied, and \Delta x is the stretch of the spring with respect to its equilibrium position. Using the data, we find
k= \frac{15 N}{3.0 cm}=5.0 N/cm

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
\Delta x =  \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm
4 0
3 years ago
Read 2 more answers
Use the image below to answer the following question (ruler not to scale).
Svetradugi [14.3K]

Answer:

it depends on wether the + and - are facing eachother

or away from eachother

Explanation:

4 0
2 years ago
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Determine the direction of the force that will act on the charge in each of the following situations. A negative charge moving t
wlad13 [49]

Answer:

a) DOWN direction,  b)  directed INTO THE SCREEN, c)    F = 0

Explanation:

The direction of the force is

for electric force

           F = q E

where we assume a positive test charge, for which the force has the direction of the electric field.

For a magnetic field

in this case the direction of the force is given by the right hand rule.

For a positive test charge, the thumb points in the direction of velocity, the other fingers extended in the direction of the magnetic field, and the palm gives the direction of force for a positive charge.

           F = q v x B

Let us apply these considerations to our case.

a) negative charge moving to the left

in a magnetic field points away from the screen

In this case the thumb goes to the left, the fingers extended outwards and the palm points upwards, but since the charge is negative the force has a DOWN direction.

b) negative charge moves to the left

in electric field it points off the screen.

The outside is in the direction of the electric field and since the charge is negative, the force is directed INTO THE SCREEN

c) positive charge moves down

in magnetic field points up

in this case the velocity and the field have the same direction so the vector product of them is zero

       F = q v  B sin 0

       F = 0

6 0
3 years ago
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