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densk [106]
2 years ago
13

Look at the illustration , what concepts of heat are being shown in this pictures ​

Physics
1 answer:
Phoenix [80]2 years ago
3 0
<h2>            (っ◔◡◔)っ ♥ Hey Honey Here<em>!</em> ♥</h2>

Answer:

♥ Convection.

♥ Conduction.

♥ Radiation.

                              ♥ (  お力になれて、嬉しいです!) ♥

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AfilCa [17]
A=f/m
Example a=10/2
A=5
7 0
3 years ago
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what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s
yaroslaw [1]

Answer : The value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

Explanation :

The expression used for second order kinetics is:

kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}

where,

k = rate constant = ?

t = time = 17s

[A_t] = final concentration = 0.0981 M

[A_o] = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

k\times 17s=\frac{1}{0.0981M}-\frac{1}{0.657M}

k=0.51M^{-1}s^{-1}

Therefore, the value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

6 0
3 years ago
A 80-kg person sits on a 2.7kg chair. Each leg of the chair makes contact with the floor on a circle that is 1.3 cm in diameter.
Rashid [163]

Answer:

need help with this 2

Explanation:

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6 0
2 years ago
Read 2 more answers
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
For states with larger values of n that admit a d subshell, the d subshell is at even higher energy than the p subshell. In fact
frosja888 [35]

Answer:

Explanation:

Correct order of energy of energy states within an atom is as follows

1s , 2 s, 2 p 3 s ,3 p, 4s , 3d,  4p, 4p,5s etc.

There is discrepancies in this order due to overlapping of outer orbitals with inner orbitals so sometimes outer orbitals have lower energy.

4 0
3 years ago
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