a load of 12tonnes is put along a horizontal plane by a force at 30°to and above the flat. if the coefficient of sliding friction is 0.2 find the frictional force
Answer:
20368.917N
Explanation:
Frictional force (F) is the product of the Coefficient of friction and the normal reaction.
F = μN
Coefficient of friction, μ = 0.2
Normal reaction = MgCosθ
Mass, m = 12 tonnes = 12 * 1000 = 12000 kg
N = 12000 * 9.8 * cos30
N = 101844.58
F = 0.2 * 101844.58
F = 20368.917N
The pressure less than atmospheric pressure is known as:
Suction pressure
Negative gauge pressure
Vacuum pressure
All of the above
Answer:
answer is option (d) all of the above
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
... is an actual sequence of interactions (i.e., an instance) describing one specific situation; a ... is a general sequence of interactions (i.e., a class) describing all possible ... associated with a situation. ... are used as examples and for clarifying details with the client. ... are used as complete descriptions to specify a user task or a set of related system features.
Answer:
ScenarioUse caseScenariosScenariosUse caseExplanation:
A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.
An insulated closed piston–cylinder device initially contains 0.3 m3 of carbon dioxide at 200 kPa
and 27°C. A resistance heater inside the cylinder is turned on and supplied heat to the gas. As a
result, the gas expanded by pushing the piston up, until the volume doubled. During this process,
6
the pressure changed according to = 4, in which the constant 6 has units of kPa.m
a) Find the mass of the hydrogen in the tank in kg.
b) Determine the work done by the gas in kJ.
To solve this problem, we can use the ideal gas law and the equation for polytropic process.
What is ideal gas law ?The ideal gas law is a fundamental law of physics that describes the behavior of an ideal gas. It relates the pressure, volume, temperature, and number of particles of a gas using the following equation:
PV = nRT
a) First, we need to find the mass of the carbon dioxide in the tank. The ideal gas law is:
PV = mRT
where P is the pressure, V is the volume, m is the mass, R is the universal gas constant, and T is the temperature. Rearranging for the mass, we get:
m = PV / RT
Substituting the given values, we have:
m = (200 kPa)(0.3 m3) / [(0.287 kPam3/kgK)(27°C + 273.15)] = 3.87 kg
So the mass of the carbon dioxide in the tank is 3.87 kg.
b) To determine the work done by the gas during the process, we can use the equation for polytropic process:
P1V1^n = P2V2^n
where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and n is the polytropic index. Substituting the given values, we have:
(200 kPa)(0.3 m3)^n = (4)(0.6 m3)^n
Dividing both sides by (0.3 m3)^n and taking the logarithm of both sides, we get:
log(200) + nlog(0.3) = log(4) + nlog(0.6)
Solving for n, we get:
n = log(4/200) / log(0.6/0.3) ≈ 1.235
Using the polytropic work equation:
W = (P2V2 - P1V1) / (1 - n)
Substituting the given values, we have:
W = [(4 kPa)(0.6 m3) - (200 kPa)(0.3 m3)] / (1 - 1.235) = 233.7 kJ
So the work done by the gas during the process is 233.7 kJ.
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The three sub regions of South America are the Andes Mountains, the Amazon Rainforest, and the Eastern Highlands. The Atacama Desert is the driest place on Earth.
Answer:
<:
Explanation:
Answer:
d
Explanation:
Find the derivative of x
Answer:
this is your answer. if mistake don't mind.
Which of the following devices is a simple machine?
A.
an engine
B.
a pulley
C.
a motor
D.
a bicycle
E.
a crane
Answer:
A PULLY
Explanation:
HAD THIS ONE THAT IS THE CORRECT ANWSER
Answer:
The answer is B. a pulley
Explanation:
I hope I answered your question:)
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:
An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter D=1 mm, electrical resistivity rhoe=10−6Ω⋅m, thermal conductivity k=25W/m⋅K, and emissivity ε=0.20. The heater is designed to deliver air at a temperature of T[infinity]=50∘C under flow conditions that provide a convection coefficient of h=250W/m2⋅K for the wire. The temperature of the housing that encloses the wire and through which the air flows is Tsur=50∘C. If the maximum allowable temperature of the wire is Tmax=1200∘C, what is the maximum allowable electric current I? If the maximum available voltage is ΔE=110V, what is the corresponding length L of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.
Solution :
Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :
[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]
[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]
[tex]$h_T=46.3 \ W/m^2 .K$[/tex]
The total heat transfer coefficient will be :
[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]
[tex]$=296.3 \ W/m^2 .K$[/tex]
Now calculating the maximum volumetric heat generation :
[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]
[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]
[tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]
The heat generation inside the wire is given as :
[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]
Here, R is the resistance of the wire
V is the volume of the wire
∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]
[tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]
where, ρ is the resistivity.
[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]
[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]
= 28.96 A
Now considering the relation for the current flow through the finite potential difference.
[tex]$E=I_{max} \times R$[/tex]
[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]
[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]
[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]
= 2.983 m
Now calculating the power rating of the heater:
[tex]$P= E \times I_{max}$[/tex]
[tex]$P= 110 \times 28.96}$[/tex]
= 3185.6 W
= 3.1856 kW
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
Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and (v) time invariant. For invertibility, either find an inverse system or an example of two inputs that lead to the same output. Note that y[n] denotes the system output and x[n] denotes the system input.
a. y[n] = x[n] x[n-1] + [n+1]
b. y[n] = cos(x[n])
Answer:
a.
y[n] = x[n] x[n-1] x[n+1]
(i) Memory-less - It is not memory-less because the given system is depend on past or future values.
(ii) Causal - It is non-casual because the present value of output depend on the future value of input.
(iii) Invertible - It is invertible and the inverse of the given system is [tex]\frac{1}{x[n] . x[n-1] x[n+1]}[/tex]
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.
b.
y[n] = cos(x[n])
(i) Memory-less - It is memory-less because the given system is not depend on past or future values.
(ii) Causal - It is casual because the present value of output does not depend on the future value of input.
(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .
For example - for x[n] = 0 , y[n] = cos(0) = 1
for x[n] = 2[tex]\pi[/tex] , y[n] = cos(2[tex]\pi[/tex]) = 1
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.
how skateboards works?
Answer
The skateboarder applies pressure to the trucks and gives/releases pressure on the levers. Second, the wheels and the axles are also examples of simple machines. They help the skater ride, spin, grind, and do a bunch of other radical movements on a skateboard.:
Explanation:
A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction. The axial load is 30 kips. Soil conditions consist of a normally consolidated clay layer with following properties; su=800 psf, andγsat= 110 pcf. Design for a FOS of 3. Assume frost depth to be 3ft below grade
Solution :
Finding the cohesion of the soil(c) using the relation:
[tex]$c = \frac{q_u}{2}$[/tex]
Here, [tex]$q_u$[/tex] is the unconfined compression strength of the soil;
[tex]$c = \frac{800}{2}$[/tex]
= 400 psf
∴ The cohesion value is greater than 0
So the use of the angle of internal friction is 0
Referring to the table relation between bearing capacity factors and angle of internal friction.
For the angle of inter friction [tex]$0^\circ$[/tex]
[tex]$N_c = 5.14$[/tex]
[tex]$N_q = 1.0$[/tex]
[tex]$N_r = 0$[/tex]
Therefore,
[tex]$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$[/tex]
= 2386 psf
∴ Allowable bearing capacity [tex]$q_{a} = \frac{Q_{allow}}{A}$[/tex]
[tex]$=\frac{30}{B^2}$[/tex]
∴ [tex]$q_a = \frac{q_{ult}}{F.O.S}$[/tex]
[tex]$\frac{30}{B^2} = \frac{2386}{3}$[/tex]
∴ B = 0.2 ft
Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft
[tex]$=0.04 \ ft^2$[/tex]
What current works best when the operator
encounters magnetic arc blow?
•DCEP
•ACEN
•CC
•AC
Answer:
AC
Explanation:
One situation when alternating current would work better than direct current is if the operator is encountering magnetic arc blow.
Current works best when the operator encounters magnetic arc blow is AC
Magnetic arc blow is simply defined as the arc deflection due to the warping of the magnetic field that is produced by electric arc current.
This is caused as a result of the following;
- if the material being welded has residual magnetism at an intolerable level
- When the weld root is being made, and the welding current is direct current which indicates constant direction and maintains constant polarity (either positive or negative).
Since it is caused by DC(Direct Current) which means constant polarity , it means the opposite will be better which is AC(alternating current) because it means that electricity direction will be switching to and fro and as such the polarity will also be revered in response to this back and forth switch manner.Thus, Current works best when the operator encounters magnetic arc blow is AC
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