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2 years ago

Which level of acceleration should you use when accelerating on a short highway entry ramp?

Engineering

1 answer:

ahrayia [7]2 years ago

5 0

The lowest level of acceleration should be use when accelerating on a short highway entry ramp.

<h3>What is an acceleration?</h3>

This means the rate at which velocity changes with time in terms of both the speed and direction.

In a machine, this means rate at which a vehicle can increase its speed in terms of the time that it takes to reach a particular speed.

Because the short highway entry ramp need careful driving, then, the lowest level of acceleration should be use when accelerating on a short highway entry ramp.

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The flexural strength or MOR of a ceramic is 310 MPa. A block of the ceramic, which is 20 mm wide, 15 mm high, and 300 mm long,

Alja [10]

Answer:

$F=6200\ \text{N}\\$

Explanation:

In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:

In this problem the flexural strength is defined with the following formula:

$\sigma=\cfrac{3FL}{2bd^2}$

where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.

The force is then defined as:

$F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}$

3 0

3 years ago

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 1.7. If, afte

luda_lava [24]

Answer:

It would take approximately 305 s to go to 99% completion

Explanation:

Given that:

y = 50% = 0.5

n = 1.7

t = 100 s

We need to first find the parameter k from the equation below.

$exp(-kt^n)=1-y$

taking the natural logarithm of both sides:

$-kt^n=ln(1-y)\\kt^n=-ln(1-y)\\k=-\frac{ln(1-y)}{t^n}$

Substituting values:

$k=-\frac{ln(1-y)}{t^n}= -\frac{ln(1-0.5)}{100^1.7} = 2.76*10^{-4}$

Also

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

Substituting values and y = 99% = 0.99

$t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s$

∴ t ≅ 305 s

It would take approximately 305 s to go to 99% completion

8 0

2 years ago

Read 2 more answers

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0

3 years ago

An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side

Aliun [14]

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to = 4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × $10^{6}$ + 4 × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × $10^{6}$ - 4 × 10³

lower side frequencies = 3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies - lower side frequencies

total width bandwidth = 8kHz

6 0

3 years ago

A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 3

aleksandrvk [35]

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to 250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>

<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook</em>

average convection heat transfer coefficient = 25.04 W/m^2 .k

<u>ii) Determine how long this process has taken </u>

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

3 0

2 years ago