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AlekseyPX
3 years ago
10

Why do we write proton ions first before electron ions? ​

Engineering
1 answer:
allsm [11]3 years ago
6 0
Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations.

Hope that hepls
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Multiply. Write the answer in simplest form. 1 3/10×1/8
kicyunya [14]

9514 1404 393

Answer:

  13/80

Explanation:

The product is ...

  (1 3/10)×(1/8) = (13/10)×(1/8) = (13×1)/(10×8) = 13/80

4 0
2 years ago
How do you make a robotic dog
fredd [130]
In order to create a robotic dog, you are needing the necessary parts to create Goddard from Jimmy nutreon boy genius
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3 years ago
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Name five or more items that were and may still be made by blacksmith
IgorLugansk [536]
Tools, weapons, hardware, armor
8 0
3 years ago
A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this
tresset_1 [31]

Answer:

a)COP=5.01

b)W_{in}=2.998 KW

c)COP=6.01

d)Q_R=17.99 KW

Explanation:

Given

T_L= -12°C,T_H=40°C

For refrigeration

  We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

COP=\dfrac{T_L}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{261}{313-261}

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that  COP=\dfrac{RE}{W_{in}}

RE is the refrigeration effect

So

5.01=\dfrac{15}{W_{in}}

b)W_{in}=2.998 KW

For heat pump

So COP of heat pump is given as follows

COP=\dfrac{T_h}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{313}{313-261}

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at  low temperature+work in put

Q_R=Q_A+W_{in}

Given that Q_A=15KW

We know that  COP=\dfrac{Q_R}{W_{in}}

COP=\dfrac{Q_R}{Q_R-Q_A}

6.01=\dfrac{Q_R}{Q_R-15}

d)Q_R=17.99 KW

5 0
3 years ago
A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the co
murzikaleks [220]

Since the applied stress required for failure due to crack propagation is still higher than 550 MPa, the ceramic is expected to fail due to overload and not because of the flaws

Explanation:

<u>Plane -Strain Fracture toughness is calculated as</u>

k_{IC}=fб\sqrt{\pi a}

F=geometry factor of the flaw

б=Stress applied

k_{IC}=Fracture toughness

a=Flaw size

<u>Given that </u>

Internal Flaw,a=0.001cm

Fracture Toughness k_{IC}=45MPa\sqrt{m}

Tensile Strength б=550 MPa

Geometry Factor,f=1

<u>Calculation</u>

An internal Flaw i s 0.001 cm

2a=0.001cm

a=0

6 0
3 years ago
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