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frutty [35]
3 years ago
8

What does the simplify command do

Engineering
1 answer:
uranmaximum [27]3 years ago
4 0

Answer:The simplify command is used to apply simplification rules to an expression. The simplify routine searches the expression for function calls, square roots, radicals, and powers and invokes the appropriate simplification procedures. For detailed information on the simplify command, see simplify/details.

Explanation:

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A rocket is launched from rest with a constant upwards acceleration of 18 m/s2. Determine its velocity after 25 seconds
lisabon 2012 [21]

Answer:

The final velocity of the rocket is 450 m/s.

Explanation:

Given;

initial velocity of the rocket, u = 0

constant upward acceleration of the rocket, a = 18 m/s²

time of motion of the rocket, t = 25 s

The final velocity of the rocket is calculated with the following kinematic equation;

v = u + at

where;

v is the final velocity of the rocket after 25 s

Substitute the given values in the equation above;

v = 0 + 18 x 25

v = 450 m/s

Therefore, the final velocity of the rocket is 450 m/s.

5 0
2 years ago
What type of bridge is the sunshine skyway bridge?
sdas [7]

Answer:

That's either a cable-stayed bridge or a cantilever bridge

6 0
3 years ago
H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate
Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0 (1)

Where:

\dot m - Mass flow, in kilograms per second.

h_{in}, h_{out} - Specific enthalpies at inlet and outlet, in kilojoules per second.

v_{in}, v_{out} - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

p = 2000\,kPa

T = 300\,^{\circ}C

h_{in} = 3024.2\,\frac{kJ}{kg}

\nu_{in} = 0.12551\,\frac{m^{3}}{kg}

Where \nu_{in} is the specific volume of water at inlet, in cubic meters per kilogram.  

Outlet (Superheated steam)

p = 600\,kPa

T = 160\,^{\circ}C

h_{out} = 2758.9\,\frac{kJ}{kg}

If we know that \dot m = 50\,\frac{kJ}{kg}, h_{in} = 3024.2\,\frac{kJ}{kg}, h_{out} = 2758.9\,\frac{kJ}{kg} and v_{in} = 30\,\frac{m}{s}, then the flow speed at outlet is:

35765-25\cdot v_{out}^{2} = 0 (2)

v_{out} \approx 37.823\,\frac{m}{s}

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius (r_{in}), in meters, by this expression:

\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}} (3)

If we know that \dot m = 50\,\frac{kJ}{kg}, v_{in} = 30\,\frac{m}{s} and \nu_{in} = 0.12551\,\frac{m^{3}}{kg}, then the inlet radius is:

r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}

r_{in}\approx 0.258\,m

The inlet radius of the nozzle is approximately 0.258 meters.  

7 0
2 years ago
A rotating beam is subject to an alternating stress of 48 kpsi and a mean stress of 24 kpsi. The ultimate strength of the materi
Alisiya [41]

Answer:

goodman = 0.694

life of beam = 211597

Explanation:

alternating stress = 48 kpsi

mean stress = 24 kpsi

ultimate strength = 100 kpsi

endurance limit = 40 kpsi

goodman:

= \frac{mean stress}{ultimate stress} +\frac{alternating stress}{endurance limit} =\frac{1}{N}

= \frac{24}{100} +\frac{48}{40} =\frac{1}{N}

= 0.24 + 1.2 = \frac{1}{N}

N = 1/1.44

N = 0.694

2. check attachment for diagram

Log(N)-3/3 = log90 - log48/log90 - log40

Log(N)-3/3 = 0.77517

Log N = 5.325509

N = 10^(5.325509)

N = 211597

6 0
3 years ago
Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e −2t + 3t
IrinaK [193]

Answer:

V=L(di/dt) where i is current, V=0.208

Explanation:

using expression iL(t)=5e-2t+3te-2t-2 and L=0.05H(50/1000)

V=0.05*d(5e-2t+3te-2t-2)/dt

since there is no power of e, I'll assume the power to be 1

V=0.05*(-2+3e-2)

at t=0.25

V=0.15e-0.2

V=0.208

3 0
3 years ago
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