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BartSMP [9]
3 years ago
6

Which elements have similar behavior

Physics
1 answer:
Nana76 [90]3 years ago
4 0

The elements which have similar behavior are Barium, strontium and beryllium.

Explanation:

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How many neutrons does an atom of Dubnium have in its nucleus
attashe74 [19]
Dubnium is a synthetic element ... Number 105.  It can be created
in a laboratory but it's not found in nature.

Fifteen (15) different isotopes of Dubnium have been created.  They all
have 105 protons in the nucleus of every atom, but they have anywhere
from 256 to 270 neutrons in each nucleus.

The most stable isotope of Dubnium is  ²⁶⁸Db , with 163 neutrons in the
nucleus.  That form of Dubnium atom has a 50-50 chance of surviving
for around 28 hours before it falls apart.
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5 0
3 years ago
A particle moving on a plane curve what is the degree of freedom
kirill115 [55]

A particle confined to move along a curved path has only one degree of freedom. inclined plane are some examples of constrained motion. Every condition of constraint reduces the number of degree of freedom by one.

I hope this helps!

7 0
3 years ago
WhAT is the change IN THE entropy of 2.0kg of h2o molecules when transform at a constant pressure of 1 atm from water at 100 deg
Fynjy0 [20]

Answer:

The entropy change is 45.2 kJ/K.

Explanation:

mass of water at 100 C = 2 kg

Latent heat of vaporization, L = 2260 kJ/kg

Heat is

H = m L

H = 2 x 2260 = 4520 kJ

Entropy is given by

S = H/T = 4520/100 = 45.2 kJ/K

3 0
2 years ago
What do neon,oxygen and nitrogen have in common?
julsineya [31]
Their Period number is common means their "Principal Quantum Numbers" are same

Hope this helps!
7 0
3 years ago
Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically t
faltersainse [42]

Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

4 0
2 years ago
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