Explain mathematically whether the divergence of the curl of a vector field is

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6

AnnZ [28]

Answer:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

Explanation:

The acceleration field is obtained by deriving the components in function of the time. That is to say:

$\frac{du}{dt}=2.05\cdot \frac{dx}{dt}+0.656\cdot \frac{dy}{dt}\\\frac{dv}{dt} = -2.18\cdot \frac{dx}{dt} -2.05\cdot \frac{dy}{dt}$

Where $\frac{dx}{dt} = u$ and $\frac{dy}{dt} = v$ .

The velocity components at given point are, respectively:

$u = 2.424$

$v=-5.266$

Lastly, the acceleration components are found:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

7 0

3 years ago

How can goal setting help with academic performance?

Zina [86]

Answer:

B.)It is an effective study skill.

Explanation:

8 0

3 years ago

How much power would you need to cool down a closed, 1 Liter container of water from 100°C to 20°C in 5 minutes? (a) 1.1W (b)1.1

qaws [65]

Answer:

The power required to cool the water is 1.11Kw.

Hence the correct option is (b).

Explanation:

Power needed to cool down is equal to heat extract from the water.

Given:

Volume of water is 1 liter.

Initial temperature is 100C.

Final temperature is 20C.

Time is 5 minutes.

Take density of water as 100 kg/m3.

Specific heat of water is 4.186 kj/kgK.

Calculation:

Step1

Mass of the water is calculated as follows:

$\rho=\frac{m}{V}$

$1000=\frac{m}{(1l)(\frac{1m^{3}}{1000l})}$

m=1kg

Step2

Amount of heat extraction is calculated as follows:

$Q=mc\bigtriangleup T$

$Q=1(4.186kj/kgk)(\frac{1000 j/kgk}{1 kj/kgk})\times(100-20)$

Q=334880 j.

Step3

Power to cool the water is calculated as follows:

$P=\frac{Q}{t}$

$P=\frac{334880}{(5min)(\frac{60s}{1min})}$

P=1116.26W

or

$P=(1116.26W)(\frac{1Kw}{1000 W})$

P=1.11 Kw.

Thus, the power required to cool the water is 1.11Kw.

Hence the correct option is (b).

8 0

4 years ago

6. At a construction site, cement, sand, and gravel are used to make concrete. The ratio of cement to sand to gravel is 1 to 2.4

S_A_V [24]

Answer:

Mass of cement used is 62.5 lb

Mass of gravel used is 225 lb

Explanation:

The ratio given here is cement to sand to gravel = 1 : 2.4 : 3.6

So, for 150 lb of sand

C : S : G = 1 : 2.4 : 3.6

$\frac{C}{S}=\frac{1}{2.4}\\\Rightarrow C=S\frac{1}{2.4}\\\Rightarrow C=150\frac{1}{2.4}\\\Rightarrow C=62.5\ lb$

Mass of cement used is 62.5 lb

$\frac{S}{G}=\frac{2.4}{3.6}\\\Rightarrow G=S\frac{3.6}{2.4}\\\Rightarrow C=150\frac{3.6}{2.4}\\\Rightarrow C=225\ lb$

Mass of gravel used is 225 lb

7 0

3 years ago

ME<br> And what you know about love

antoniya [11.8K]

Answer:

Everything I got what you need

7 0

3 years ago