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katen-ka-za [31]
3 years ago
9

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

Engineering
1 answer:
lisabon 2012 [21]3 years ago
5 0

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.

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The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm
Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0
3 years ago
What is the single largest contributor to increasing takeoff or landing distance?
hodyreva [135]

An uphill slope improves the take-off ground run, and a downhill slope increases the landing ground run.

<h3>What is uphill slope?</h3>

Driving uphill suggests climbing a “positive” six percent slope . Driving downhill, the “rise” exists as a  drop, so there is a “negative,” or downhill, slope (Figure B). When dealing with slope, a positive slope simply indicates uphill and a negative slope indicates downhill. An uphill slope improves the take-off ground run, and a downhill slope increases the landing ground run.

If something or someone lives moving downhill or is downhill, they exist moving down a slope or are located toward the bottom of a hill. He headed downhill toward the river. adverb. If you communicate that something exists going downhill, you mean that it is becoming worse or less prosperous.

To learn more about uphill slope refer to:

brainly.com/question/13361896

#SPJ4

3 0
2 years ago
Can you determine the critical distance along a flat surface?
Keith_Richards [23]

Explanation:

Consider a fluid of density, ρ moving with a velocity, U over a flat plate of length, L.

Let the Kinematic viscosity of the fluid be ν.

Let the flow over the fluid be laminar for a distance x from the leading edge.

Now this distance is called the critical distance.

Therefore, for a laminar flow, the critical distance can be defined as the distance from the leading edge of the plate where the Reynolds number is equal to 5 x 10^{5}

And Reynolds number is a dimensionless number which determines whether a flow is laminar or turbulent.  

Mathematically, we can write,

    Re = \frac{\rho .U.x}{\mu }

or 5 x 10^{5} =  \frac{\rho .U.x}{\mu } ( for a laminar flow )

Therefore, critical distance

x=\frac{5\times 10^{5}\times \mu }{\rho \times U}

So x is defined as the critical distance upto which the flow is laminar.

6 0
3 years ago
For some metal alloy it is known that the kinetics of recrystallization obey the Avrami equation, and that the value of k in the
JulsSmile [24]

Answer:

t = 1456.8 sec

Explanation:

given data:

contant k = 2.60*10^{-6}

rate of crystallization is 0.0013 s-1

rate of transformation is given by

r = \frac{1}{t_0.5}

use specifies value to solve t_0.5

it is ime required for 50% tranformation

r = \frac{1}{.0013}=769.2 sec

Avrami equation is given by

y = 1 - e^{-kt^n}

0.5 = 1 - e^{-kt_0.5^n}

1-0.5 = e^{-kt_0.5^n}

ln (1 - 0.5) = -kt_0.5^n

ln \frac{ln (1 - 0.5)}{-k} = nln t_0.5

n = \frac{ ln \frac{ln (1 - 0.5)}{-k}}{ln t_0.5}

n = \frac{ ln \frac{ln (1 - 0.5)}{-2.60*10^{-6}}}{ln 769.2}

n = 1.88

second degree of recrystalization may be determine by rearranging original avrami equation

t = [\frac{-ln(1-y)}{k}]^{1/n}

for 90%completion

t = [\frac{-ln(1-0.9)}{2.60*10^{-6}}]^{1/1.88}

t = 1456.8 sec

5 0
3 years ago
A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middl
andre [41]

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

<u>Determine the maximum gage pressure in the tank</u>

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

6 0
3 years ago
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