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zhuklara [117]
3 years ago
6

You can mow an average of 1400 square meters each hour. How many minutes will it take you to mow a lawn with an area of 320000 s

quare feet?
Physics
1 answer:
Ksivusya [100]3 years ago
7 0

Answer:

time required to mow lawn is 1274.06 minutes

Explanation:

given data

average mow = 1400 square meters each hour

area = 320000 square feet

to find out

How many minutes will take to mow lawn

solution

we know that here 1 square feet is equal to 0.092903 square meter

so 320000 square feet will be = 0.09203 × 320000 = 29728.9728 square meter

so time required is express as

time required = \frac{distance}{speed}

time required = \frac{29728.9728}{1400}

time required = 21.23 hours

so time required = 21.23 × 60 min = 1274.06

time required to mow lawn is 1274.06 minutes

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Air flows through a nozzle at a steady rate. At the inlet the density is 2.21 kg/m3 and the velocity is 20 m/s. At the exit, the
Vitek1552 [10]

To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.

The general formula is defined by

\dot{m}=\rho A V

Where,

\dot{m} = mass flow rate

\rho = Density

V = Velocity

Our values are divided by inlet(1) and outlet(2) by

\rho_1 = 2.21kg/m^3

V_1 = 20m/s

A_1 = 60*10^{-4}m^2

\rho_2 = 0.762kg/m^3

V_2 = 160m/s

PART A) Applying the flow equation we have to

\dot{m} = \rho_1 A_1 V_1

\dot{m} = (2.21)(60*10^{-4})(20)

\dot{m} = 0.2652kg/s

PART B) For the exit area we need to arrange the equation in function of Area, that is

A_2 = \frac{\dot{m}}{\rho_2 V_2}

A_2 = \frac{0.2652}{(0.762)(160)}

A_2 = 2.175*10^{-3}m^2

Therefore the Area at the end is 21.75cm^2

3 0
3 years ago
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

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Answer:

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