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3 years ago

What affect results when there is an impact between earth and is astroids check all that apply.

Engineering

1 answer:

Vadim26 [7]3 years ago

6 0

Answer:

Massive destruction

Explanation:

If the asteroid collides with the ground, a massive volume of dust will be blasted into the environment. If it collides with water, the amount of water vapour in the atmosphere will rise. This would result in more rain, which would cause earthquakes and mudslides.

As the asteroid collided with the Earth, massive volumes of dust were ejected into to the atmosphere. The sun's light were stopped from entering the Earth's surface, which is terrible news for plants.

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Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kin

hammer [34]

Answer:

a. 149.74 KJ/KG

b. 97.9%

c. 0.81 kJ/kg K

Explanation:

8 0

4 years ago

Ronny has a hydraulic jack. The input force is 250 N, while the output force is 7,500 N. If the area of the pipe below the input

Ganezh [65]

Answer:

6 m²

Explanation:

application of fluid pressure according to Pascal's principle for the two pistons is given as:

$P_1=P_2$

Where P₁ is the pressure at the input and P₂ is the pressure at the output.

But P₁ = F₁ / A₁ and P₂ = F₂ / A₂

Where F₁ and F₂ are the forces applied at the input and output respectively and A₁ and A₂ are the area of the input pipe and output pipe respectively

Since, $P_1=P_2$

$\frac{F_1}{A_1} =\frac{F_2}{A_2}\\$

But A₁ = 0.2 m², F₁ = 250 N, F₂ = 7500 N. Substituting values to get:

$\frac{F_1}{A_1} =\frac{F_2}{A_2}\\\frac{250}{0.2}=\frac{7500}{A_2}\\ A_2=\frac{7500*0.2}{250} = 6m^2$

Therefore, the area of the pipe below the load is 6 m²

6 0

3 years ago

A 0.4-W cylindrical electronic component with diameter 0.3 cm and length 1.8 cm and mounted on a circuit board is cooled by air

Katyanochek1 [597]

Answer:

The surface temperature of the component 54.6 degrees celsius.

Explanation:

Please see attachment.

7 0

3 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago

A student wants to restate some ideas she found in a journal article by a prominent expert in economics. She combines her own wo

kipiarov [429]

Explanation:

Althought she referenced the article at the end, is imposible to know which part of the article is hers and which part is the expert's so that would be plagiarism.

If she used quotation marks in the words of the expert it would be clear and no plagiarism could be accused.

4 0

3 years ago