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AnnZ [28]
2 years ago
8

A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a

tmospheric pressure is 100 kPa, determine the final temperature of the gas in °C.
Physics
1 answer:
aleksley [76]2 years ago
5 0

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

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A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0
2 years ago
Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving
Rzqust [24]

Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

4 0
2 years ago
A bucket of water experiencing a gravitational force of 525 N is pulled from a water well. Net force in the Y direction is 45 N
vivado [14]

Answer:

T = 570 N

Explanation:

Given that,

The gravitational force acting on a bucket of water = 525 N

Net force in the Y direction is 45 N

We need to find the magnitude of the force of tension. It can be calculated as :

45 = T - 525

T = 525 + 45

T = 570 N

Hence, the force of tension is 570 N.

7 0
2 years ago
Suppose that air resistance cannot be ignored. For the position at which the person has jumped from the platform and the cord re
Kamila [148]

Answer:

The stretch cord stores potential energy as a result of stretching but due to kinetic energy, it will move back to its original state. Since air resistance is not being ignored in this case, it will experience a slight delay in stretching at first.

Explanation:

In case, where air resistance is being ignored the stretch cord will stretch as it normally does.              

  • Air resistance is a force that any object experiences as a result of its motion through the air.  

There are various factors that affect air resistance like speed, the density of air, area, the shape of an object etc. Meanwhile, the density of air changes with temperature or altitude. <em>Hence this force is not constant but is thought to be constant during short time frames.  </em>

6 0
2 years ago
Lucy has three sources of sound that produce pure tones with wavelengths of 60cm, 100cm, and 124cm.
denis23 [38]

Answer:

a) We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength

b) There is resonance for the lengths 25 and 75 cm

c) Resonance occurs for tubes with length 31 and 93 cm

Explanation:

To find the length of the tube that has resonance we must find the natural frequencies of the tubes, for this at the point that the tube is closed we have a node and the open point we have a belly; in this case the fundamental wave is

              λ = 4L

The next resonance called first harmonic    λ₃ = 4L / 3

The next fifth harmonic resonance               λ₅ = 4L / 5,

WE see that the general form is                    λ ₙ= 4L / n          n = 1, 3, 5 ...

Let's use these expressions for our problem

Let's start with the shortest wavelength.

a) Lam = 60 cm

Let's look for the tube length that this harmonica gives

               L = λ n / 4

To find the shortest tube length n = 1

               L = 60 1/4

              L = 15 cm

For n = 3

              L = 60 3/4

              L = 45 cm

For n = 5

              L = 60 5/4

              L = 75 cm

For n = 7

             L = 60 7/4

             L = 105cm

We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength, in different harmonics 1, 3 and 5

.b) λ = 100 cm

For n = 1

         L = 100 1/4

        L = 25 cm

For n = 3

        L = 100 3/4

       L = 75 cm

For n = 5

       L = 100 5/4

      L = 125 cm

There is resonance for the lengths 25 and 75 cm in the fundamental and third ammonium frequency

c) λ=  124 cm

       L = 124 1/4

       L = 31 cm

For the second resonance

      L = 124 3/4

      L = 93 cm

Resonance occurs for tubes with length 31 and 93 cm in the fundamental harmonics and third harmonics

8 0
2 years ago
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