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Ludmilka
3 months ago
9

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas i

s kept constant?"
Physics
1 answer:
Softa [3K]3 months ago
5 0

Esta pregunta está incompleta, la pregunta completa es;

La figura muestra un recipiente sellado en la parte superior por un pistón móvil. Dentro del recipiente hay un gas ideal a 1.00 atm. 20.0°C y 1.00 L.

"¿Cuál será la presión dentro del recipiente si el pistón se mueve hasta la marca de 1.60 L mientras se mantiene constante la temperatura del gas?"

Answer:

la presión dentro del recipiente será 0.625 atm si el pistón se mueve hasta la marca de 1.60 L manteniendo constante la temperatura del gas

Explicación:

Dado que;

P₁ = 1.00 atm

P₂ =?

V₁ = 1 L

V₂ = 1.60 L

la temperatura del gas se mantiene constante

sabemos que;

P₁V₁ = P₂V₂

por lo que hacemos la sustitución

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Por lo tanto, la presión dentro del recipiente será 0.625 atm si el pistón se mueve hasta la marca de 1.60 L mientras la temperatura del gas se mantiene constante

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A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 cans, 0.355 - l each,
Keith_Richards [3271]
Flow rate calculations yield 220 cans, each with a volume of 0.355 l, leading to 78.1 l/min or 1.3 l/s or 0.0013 m³/s.

At Point 2:
A2 = 8 cm² = 0.0008 m²
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

At Point 1:
A1 = 2 cm² = 0.0002 m²
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 =?
Height = 1.35 m

Using Bernoulli’s principle;
P2 + 1/2 * V2² / density = P1 + 1/2 * V1² / density + density * gravitational acceleration * height
=> 152000 + 0.5 * (1.625)² * 1000 = P1 + 0.5 * (6.5)² * 1000 + (1000 * 9.81 * 1.35)
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31 - 34368.5 = 118951.81 Pa = 118.95 kPa
3 0
3 months ago
Read 2 more answers
You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly
Sav [3153]

A bathroom scale operates under gravitational influence. Typically, a reading is captured when your body applies force onto the scale. Yet in this scenario, as both you and the scale move downwards, your body ceases to press against the scale. Consequently, the result is:

<span>The scale reading will instantly drop to zero</span>

7 0
3 months ago
Read 2 more answers
A string is stretched by two equal but opposite forces f newton each what is tension in string
Maru [3345]

The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.

Explanation:

If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.

When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.

Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.

8 0
3 months ago
A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which
ValentinkaMS [3465]

Answer:

h = 10 m

Explanation:

Given,

mass of the platform = 50 Kg

Kinetic energy = 5000 J

height from which the diver dove =?

Taking the acceleration due to gravity as 10 m/s²

Using the conservation of energy principle

Kinetic energy is transformed into potential energy

K.E = P.E

K.E = m g h

5000 = 50 x 10 x h

500 h = 5000

h = \dfrac{5000}{500}

    h = 10 m

The height from which the diver dove is equal to h = 10 m

6 0
2 months ago
A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At
kicyunya [3294]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + \frac{1}{2}at^{2}

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + \frac{1}{2}(2.25 x 15.4^{2})

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

v^{2} = u^{2} + 2gs

0^{2} = 34.65^{2} + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying v^{2} = u^{2} + 2gs

v^{2} = 0^{2} + 2(9.81 x 328)

v^{2} = 0 + 6435.36

v = \sqrt{6435.36} = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0
3 months ago
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