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2 years ago

Which of the following fluid power systems is portable and human-scale?

Engineering

2 answers:

meriva2 years ago

7 0

Answer:

a fluid power engine okk done please

faltersainse [42]2 years ago

3 0

Answer:

A fluid powered drill

Explanation:

Firefighters use these at car accidents because of their durability reliability and incredible torque

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A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end

hjlf

Answer:

a) $v=19.6 m/s$

b) $H=19.58 m$

c) $v_{f}=29.57 m/s$

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

$W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J$

But we know the work is equal to change of kinetic energy, so:

$W=\Delta K=\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2W}{m}}=19.6 m/s$

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

$v_{f}^{2}=v_{i}^{2}-2gh$

At the maximum height the velocity is 0, so v(f) = 0.

$0=v_{i}^{2}-2gH$

$H=\frac{19.6^{2}}{2*9.81}=19.58 m$

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

$v_{f}^{2}=v_{i}^{2}-2gh$

$v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s$

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0

3 years ago

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Nataly_w [17]

Outdoor advertising

8 0

4 years ago

Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i

DedPeter [7]

Answer:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

Explanation:

The internal energy is equal to:

$dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV$

The entropy is equal to:

$dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV$

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

$\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }$

Replacing:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

4 0

3 years ago

The pressure and temperature at the beginning of compression of a cold air-standard Diesel cycle are 100 kPa and 300K, respectiv

hram777 [196]

Answer:

Cut off ratio=2.38

Explanation:

Given that

$T_1=300K$

$P_1=100KPa$

$P_2=P_3=7200KPa$

$T_3=2250K$

Lets take $T_1$ is the temperature at the end of compression process

For air γ=1.4

$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}$

$\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}$

$T_2=1070K$

At constant pressure

$\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}$

$\dfrac{2550}{1070}=\dfrac{V_3}{V_2}$

$\dfrac{V_3}{V_2}=2.83$

So cut off ratio

$cut\ off\ ratio =\dfrac{V_3}{V_2}$

Cut off ratio=2.38

6 0

3 years ago

The natural material in a borrow pit has a mass unit weight of 110.0 pcf and a water content of 6%, and the specific gravity of

enot [183]

Answer:

A. 288,030.91 cy

B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water

Explanation:

The natural material in the barrow properties are;

The mass unit weight, γ = 110.0 pcf

The water content, w = 6%

The specific gravity of the soil solids, $G_s$ = 2.63

The desired dry unit weight, $\gamma _d$ = 122 pcf

The water content, w₁ = 5.5 %

The net section volume, $V_T$ = 245,000 cy = 6,615,000 ft³

A. $\gamma _d$ = $W_s$ / $V_T$

∴ $W_s$ = $V_T$ × $\gamma _d$ = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs

w = ( $W_w$ / $W_s$ ) × 100

∴ $W_w$ = (w/100) × $W_s$ = (6/100) × 807030000 lbs = 48421800 lbs

The weight of solids

W = $W_s$ + $W_w$ = 807030000 lbs + 48421800 lbs = 855451800 lbs

V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy

V = 288,030.91 cy

The amount of cubic yards of borrow required = 288,030.91 cy

B. The volume of water in the required soil is found as follows;

$W_{w1}$ = (w₁/100) × $W_s$ = (5.5/100) × 807030000 lbs = 44386650 lbs

The amount of water that must be added = $W_{w1}$ - $W_w$ = 44386650 lbs - 48421800 lbs = -4,035,150 lbs

Therefore, 4,035,150 lbs of water must be removed

The density of water, ρ = 8.345 lbs/gal

Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal of water must be removed from the natural material

7 0

3 years ago