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steposvetlana [31]
3 years ago
9

Explain why when water is at the boiling process in a pot it represents convection instead of conduction?​

Physics
1 answer:
Anuta_ua [19.1K]3 years ago
8 0

Answer:

When boiling water, the temperature of molecules within the water increases    

Explanation: and they slowly begin to move at a rapid rate, upwards. ... The hot water molecules become less dense, and they rise above the denser cooler molecules. This movement of molecules creates convection currents.

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A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo
BARSIC [14]

We are given that,

\frac{dx}{dt} = 4ft/s

We need to find \frac{d\theta}{dt} when x=8ft

The equation that relates x and \theta can be written as,

\frac{x}{6} tan\theta

x = 6tan\theta

Differentiating each side with respect to t, we get,

\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}

\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}

\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}

Replacing the value of the velocity

\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2

\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta

The value of cos \theta could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

cos\theta = \frac{6}{10}

\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2

\frac{d\theta}{dt} = \frac{24}{25}

Search light is rotating at a rate of 0.96rad/s

4 0
3 years ago
A 1750kg bumpercar moving at 1.50m/s to the right collides elastically with a 1450kg car going to the left at 1.10m/s. The 1750k
damaskus [11]
1984.08 kg that’s the answer
6 0
3 years ago
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
What is the approximate mass of air in a living room 4.5m×3.4m×2.9m? the density of air is 1.29 kg/m3?
topjm [15]
First we have to calculate the volume of the living room:
V = L x W x H = 4.5 m * 3.4 m * 2.9 m
V = 44.37 m³
We know that Density = 1.29 kg/m²
D = m / V
m = D · V
m = 1.29 kg/m³ · 44.37 m³
m = 57.2373 kg ≈ 57.2 kg
Answer: The approximate mass of air in living room is 57.2 kg.
6 0
3 years ago
Write SHORT examples of Gravitational potential energy and Elastic energy
Nikolay [14]
Dropping a bouncy ball and stretching a rubber ban.
6 0
3 years ago
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