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3 years ago

When mass is the same, what is the relationship between radius and compression strength?

Engineering

1 answer:

VladimirAG [237]3 years ago

4 0

Answer:

a gas in a rigid container has a pressure of 632 torrs and a temperature of 45 celsius. The pressure has increased to 842 torrs. What is the new temperature of the gas

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What is the most likely cause of a double precipitin ring in a RID test

Vladimir [108]

Answer:

The most likely cause of a double precipitin ring in a RID test is when there is an antigen-antibody reaction ratio within the equivalence zone. The precipitin line forms when the antigen (with one or more epitopes) and antibody are in optimal proportions to each other and form an insoluble precipitates.

4 0

3 years ago

The explosion of a hydrogen bomb can be approximated by a fireball with a temperature of 7200 K, according to a report published

Iteru [2.4K]

Answer:

The rate of radiation of the energy is $E_r = 1.523747635*10^9 W/m^2$

The irradiation is $G =46.177\ kW/m^2$

The amount of energy absorbed is $E_B = 461.772 KJ$

The oak Tree would catch fire because the temperature of the blast(7200 K) is higher than the flammability limit (650 K) of the oak tree and secondly the thickness is very small

Explanation:

From the question we are told that

The temperature is $T = 7200K$

The diameter of the ball is $d = 1.5 km = 1.5 *1000 = 1500m$

Hence the radius = $= \frac{1500}{2} = 750m$

The total energy radiated can be mathematically represented as

$E = \sigma A T^4$

Where $\sigma$ is the Stefan-Boltzmann constant $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}$

A is the area of a sphere = $\pi d^2 = 3.142 * 1500^2 = 7.069500 *10 ^6\ m^2$

Substituting values we have

$E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4$

$=1.077*10^{15} W$

Now the state of the energy is mathematically represented as

$Rate \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4$

$= 5.67*10^{-8} * 7200^2$

$= 1.523747635*10^9 W/m^2$

A sketch illustrating the b part of the question is shown on the first uploaded image

looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

from the diagram x can be calculated as follows

$x = \sqrt{40^2 + 16^2}$

$= 43.0813 Km$

This value of x represents the radius of the blast(assuming it is spherical ) when it is at that wall

Now the irradiation G is mathematically represented as

$G = \frac{E}{4 \pi r^2}$

Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

$G= \frac{1.077*10^15}{4 \pi (431081.3^2)}$

$G =46.177\ kW/m^2$

Generally the amount of energy absorbed can be mathematically represented as

$Amount \ of \ energy \ absorbed \ (E_B ) = G * t$

Where t is the time taken

Therefore $E_B = 46.177 *10 = 461.77 KJ$

6 0

3 years ago

Consider a venture with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir.

dangina [55]

Answer:

1913meter per second square.

Explanation:

From the Context the vacuum can be said be the presence of 5e difference below the outside ambient temperature.

For the Venturi.

Please go through the attached file for the rest of the solutions and the answer.

3 0

3 years ago

Health Maintenance Organizations (HMOs) are the new model for providing medical care. Criticisms of HMOs include all of the foll

Tema [17]

Answer:

c. multiple trips to many specialists.

Explanation:

These are a organizations that provide health insurance time fee.

7 0

4 years ago

a. Replacing standard incandescent lightbulbs with energy-efficient compact fluorescent lightbulbs can save a lot of energy. Cal

monitta

Answer:

a) 1512000 Joules

b) 5040 seconds = 84 minutes = 1.4 hours

Explanation:

a) Power saved by replacing bulbs = 60-18 = 42 W = 42 J/s

Time the bulb is used for = 10 hours

Energy saved during this time

42×10×60×60 = 1512000 Joules

Saved energy by replacing standard incandescent lightbulbs with energy-efficient compact fluorescent lightbulbs in 10 hours is 1512000 Joules

b) Power the plasma TV uses = 300 W = J/s

$\frac{1512000}{300}=5040\ s$

Time a plasma TV can be used for with the saved energy is 5040 seconds = 84 minutes = 1.4 hours.

4 0

3 years ago