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kipiarov [429]
3 years ago
11

During the collision, is the magnitude of the force of asteroid A on asteroid B greater than, less than, or equal to the magnitu

de of the force of asteroid B on asteroid A?
Engineering
2 answers:
Natasha2012 [34]3 years ago
5 0

Answer: The magnitude of the fore of asteroid A on asteroid B is equal to the magnitude of the force of asteroid B on asteroid A.

Explanation:

The force of one asteroid on the other, is explained by the Universal Law of gravitation, that can be written as follows:

Fg = G* mA*mB / (rAB)²

Even though the force is a vector, and is directed towards one of the masses, along the line that joins their centers, it can be seen the magnitude is the same.

This can be explained also stating that these forces form a pair of action-reaction forces, which, according with Newton’s 3rd Law, must be equal and opposite each other.

goblinko [34]3 years ago
3 0

Answer:

B

Explanation:

greater than, less than, or equal to the magnitude of the force of asteroid!

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With reference to the NSPE Code of Ethics, which one of the following statements is true regarding the ethical obligations of th
34kurt

Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.

Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.  

One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.  

4 0
3 years ago
Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe
Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

We know viscosity of water is\left (\mu\right )=8.90\times10^{-4}pa-s

Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

\frac{\Delta P}{L}=0.1898 Pa/m

4 0
3 years ago
- Viscoelastic stress relaxation
My name is Ann [436]

Explanation:

The correct answers to the fill in the blanks would be;

1. Viscoelastic stress relaxation refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).

2. Viscoelastic creep refers to scenarios for which a polymer will permanently flow over time in response a constant applied stress.

The polymer whose properties have been mentioned above is commonly known as Kevlar.

It is mostly used in high-strength fabrics and its properties are because of several hydrogen bonds between polymer molecules.

5 0
3 years ago
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of
Alexxandr [17]

Answer:

P_m_i_n = 442KPA

Explanation:

We are given:

m = 1.06Kg

T_H = 1.2T_L

T = 22kj

Therefore we need to find coefficient performance or the cycle

COP_R = \frac {1}{(T_R/T_l) -1}

= \frac {1 }{1.2-1}

= 5

For the amount of heat absorbed:

Q_l = COP_R Wm

= 5 × 22 = 110KJ

For the amount of heat rejected:

Q_H = Q_L + W_m

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= = \frac{132}{1.06}

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

T_L = \frac{T_H}{1.2};

T_L = \frac{342.5}{1.2}

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at T_L = 12.4°c we have 442KPa

6 0
3 years ago
Not sure which one....
Airida [17]
I think downwards as that's how most saw's work.
4 0
3 years ago
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