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choli [55]
1 year ago
10

a piston-cylinder apparatus has a 4 kg piston with a diameter of 5 cm resting on a body of water 60 cm high. atmospheric pressur

e is 101 kpa, and the temperature of the water is 250˚c. what is the mass of water in the container?
Physics
1 answer:
Leokris [45]1 year ago
8 0

The mass of the water in the container is 10,104.00 kg.

A schematic diagram of a piston-cylinder apparatus is shown in the adjoining image.

In the above case, the atmospheric pressure applied by the piston and the weight of the piston should balance the weight of the water inside the container

Hence,

Atmospheric Pressure x Area of Piston + Mass of piston x g = Mass of water x g

101 x 10^3 x pi x (2.5 x 10^-2)^2 + 4 x 10 = Mass of water x 10

Mass of water x 10 = 101,040.001

Mass of Water = 101,040.001/10

Mass of Water = 10,104 kg

Hence, the mass of the water in the container is 10,104.00 kg.

To know more about the "piston-cylinder apparatus", refer to the link given below:

brainly.com/question/13999257?referrer=searchResults

#SPJ4

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Answer:

90 Minutes

Explanation:

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t = 75 / 50

t = 1.5 hour

t = 90 minutes.

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The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v
Arturiano [62]

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position =  5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

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<u>a = 1.72 m/s²</u>

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3 years ago
How does the location of the outer planets explain their composition and characteristics?
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6 0
3 years ago
Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o
adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

F = \sqrt{F_h^2+F_v^2}

F = \sqrt{3.36^2+13.98^2}

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

\theta =tan^{-1}(\dfrac{F_v}{F_h})

\theta =tan^{-1}(\dfrac{13.98}{3.36})

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