Answer:
d.
Explanation:
Tech A says that LED brake lights illuminate faster than incandescent bulbs. Tech B says that LED brake lights have
more visibility and last longer. Who is correct?
Answer:
Both
Explanation:
Which gas is released in the SMAW process causing a
shielding affect on the molten weld pool?
•nitrogen
•carbon dioxide
•argon
•hydrogen
Now, you get a turn to practice writing a short program in Scratch. Try to re-create the program that was shown that turns the sprite in a circle. After you have completed that activity, see if you can make one of the improvements suggested. For example, you can try adding a sound. If you run into problems, think about some of the creative problem-solving techniques that were discussed.
When complete, briefly comment on challenges or breakthroughs you encountered while completing the guided practice activity.
Pls help im giving 100 points for this i have this due in minutes
Answer:
u need to plan it out
Explanation:
u need to plan it out
Answer:
use the turn 1 degrees option and put a repeat loop on it
Explanation:
u can add sound in ur loop
Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligible medium resistance. The cake solids (dry basis) per volume of filtrate was 20 g/liter. It is desired to operate a larger rotary vacuum filter (diameter 8 m and length 12 m) at a vacuum pressure of 80 kPA with a cake formation time of 20 s and a cycle time of 60 s. Determine the filtration rate in volumes/hr expected for the rotary vacuum filter.
Answer:
5.118 m^3/hr
Explanation:
Given data:
viscosity of cell broth = 5cP
cake resistance = 1*1011 cm/g
dry basis per volume of filtrate = 20 g/liter
Diameter = 8m , Length = 12m
vacuum pressure = 80 kpa
cake formation time = 20 s
cycle time = 60 s
Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter
attached below is a detailed solution of the question
Hence The filtration rate in volumes/hr expected for the rotary vacuum filter
V' = ( [tex]\frac{60}{20}[/tex] ) * 1706.0670
= 5118.201 liters ≈ 5.118 m^3/hr
The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.
Answer:
19.81 m/s
Explanation:
The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:
[tex]a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=\frac{v^2}{r},a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=\frac{v^2}{r} \\\\v^2=ar\\\\v=\sqrt{ar} \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=\sqrt{0.4*9.81*100} \\\\v=19.81\ m/s[/tex]
1. What is the productivity rate using cycle time for the following information:
I
Type of Work – Hauling
Average Cycle Time – 35 Minutes
Truck Capacity – 25 Tons
Crew - One Driver
Productivity Factor - 0.85
System Efficiency – 55 Minutes
per
Hour
A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN uniformly distributed compressive force in the direction 2, which is perpendicular to the fiber direction (direction 1). The cube is constrained against expansion in direction 3. Determine:
a. changes in the 50-mm dimensions.
b. stresses required to provide constraints.
Answer:
hello some parts of your question is missing attached below is the missing part
answer :
A) Determine changes in the 50-mm dimensions
The changes are : 0.006mm compression in y-direction
0.002 mm expansion in x and z directions
B) the stress required are evenly distributed
Explanation:
Given data :
50-mm cube of graphite fiber reinforced polymer matrix
subjected to 125-KN force in direction 2,
direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3
A) Determine changes in the 50-mm dimensions
The changes are : 0.006mm compression in y-direction
0.002 mm expansion in x and z directions
B) the stress required are evenly distributed
attached below is the detailed solution
A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16.5 s later. The descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude. Assume that g = 32.2 ft/s2.
Determine
(a) the speed v1 of the rocket at the end of powered flight,
(b) the maximum altitude reached by the rocket.
Answer:
[tex]u = 260.22m/s[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Explanation:
Given
[tex]S_0 = 89.6ft[/tex] --- Initial altitude
[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds
[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)
Solving (a): Final Speed of the rocket
To do this, we make use of:
[tex]S = ut + \frac{1}{2}at^2[/tex]
The final altitude after 16.5 seconds is represented as:
[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]
Substitute the following values:
[tex]S_0 = 89.6ft[/tex] [tex]S_{16.5} = 0ft[/tex] [tex]a = -g = -32.2ft/s^2[/tex] and [tex]t = 16.5[/tex]
So, we have:
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]
[tex]0 = 89.6 + 16.5u- 4383.225[/tex]
Collect Like Terms
[tex]16.5u = -89.6 +4383.225[/tex]
[tex]16.5u = 4293.625[/tex]
Make u the subject
[tex]u = \frac{4293.625}{16.5}[/tex]
[tex]u = 260.21969697[/tex]
[tex]u = 260.22m/s[/tex]
Solving (b): The maximum height attained
First, we calculate the time taken to attain the maximum height.
Using:
[tex]v=u + at[/tex]
At the maximum height:
[tex]v =0[/tex] --- The final velocity
[tex]u = 260.22m/s[/tex]
[tex]a = -g = -32.2ft/s^2[/tex]
So, we have:
[tex]0 = 260.22 - 32.2t[/tex]
Collect Like Terms
[tex]32.2t = 260.22[/tex]
Make t the subject
[tex]t = \frac{260.22}{ 32.2}[/tex]
[tex]t = 8.08s[/tex]
The maximum height is then calculated as:
[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]
This gives:
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]
[tex]S_{max} = 1141.0676[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Hence, the maximum height is 1141.07ft
Find the derivative of x
Answer:
this is your answer. if mistake don't mind.
I don’t know the answer to this question
Answer:
I dont know the answer either
Explanation:
Answer:
flux
Explanation:
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 2.00 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.
Answer:
Explanation:
From the given information:
The equation for applied stress can be expressed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]
where;
[tex]\phi[/tex] = angle between the applied stress [100] and [111]
To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system
Using the equation:
[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [111]
[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]
Thus;
[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]
[tex]\phi= 54.74^0[/tex]
To determine [tex]\lambda[/tex] for [tex][1 \overline 1 0][/tex]
where;
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [tex][1 \overline 1 0][/tex]
[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]
Thus;
[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]
[tex]\phi= 45^0[/tex]
Thus, the magnitude of the applied stress can be computed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]
[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]
[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]