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3 years ago

9

A particle moves along a straight line with a velocity V=(200s) mm/s, where s is in millimeters. Determine acceleration of the p

article at S = 2000mm.

1 answer:

iragen [17]3 years ago

8 0

Answer:

200 mm/s²

Explanation:

See it in the pic

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What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit

jeka94

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

$v=\dfrac{K}{n_e}\dfrac{dh}{dl}$

$v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s$

$v=3.22\times 10^{-6}\ m/s$

The velocity in m/d ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

4 0

3 years ago

Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz

butalik [34]

Answer:

Exit velocity $V_2=1472.2$ m/s.

Explanation:

Given:

At inlet:

$P_1=100 bar,T_=600°C,V_1=35m/s$

Properties of steam at 100 bar and 600°C

$h_1=3624.7\frac{KJ}{Kg}$

At exit:Lets take exit velocity $V_2$

We know that if we know only one property inside the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

$h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}$

So the enthalpy of steam at the exit of turbine

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}$

$h_2=2541.62\frac{KJ}{Kg}$

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

In the case of adiabatic nozzle Q=0,W=0

$3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0$

$V_2=1472.2$ m/s

So Exit velocity $V_2=1472.2$ m/s.

4 0

3 years ago

Between chain, belt and gear drive, which causes more friction?

loris [4]

Answer:

All of them cause friction

Explanation:

google

6 0

3 years ago

A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s

monitta

Answer:

a. $\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$ which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

$v_1 = \dfrac{4}{3} \times \pi \times r^3$

$\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3$

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

$\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}$

We have;

$\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$

$\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )$

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

$\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}$

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. $W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}$

$p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }$

p₂ = 100000/0.941 = 106.265 kPa

$W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J$

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

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3 years ago

Is it true or false that sometimes people except certain risks in exchange for other benefits

mixas84 [53]

Yes, they have to make certain sacrifices ! for many many benefits.

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3 years ago