An escalator with 35° incline is designed to have two passengers per step. Find number of persons moved per hour for the design if velocity is 50cm/mins and the step tread is 600mm.
Answers
Answer:
The escalator disposes 58 passengers each hour.
Explanation:
The velocity diagram of the escalator is shown in the attached figure
We can obtain the vertical distance that the escalator moves in 1 hour as
Thus in 1 hour the last thread moves 1720.73 cm
Now it is given that 1 thread = 600 mm =60 cm
Thus the number of times the last thread moves equals
Since each time last thread moves it disposes 2 passengers thus the number of passengers disposed when the thread moves 28.68 times equals
Thus the escalator disposes 58 passengers each hour.
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An inventor claims that he wants to build a dam to produce hydroelectric power. He correctly realizes that civilization uses a lot more electricity during the day than at night, so he thinks he has stumbled upon a great untapped energy supply. His plan is to install pumps at the bottom of the dam so that he can pump some of the water that flows out from the generators back up into the reservoir using the excess electricity generated at night. He reasons that if he did that, the water would just flow right back down through the generators the next day producing power for free. What is wrong with his plan?
Cody’s car accelerates from 0m/s to 45 m/s northward in 15 seconds. What is the acceleration of the car 
Answers
Answer:
(b) False
Explanation:
Isothermal process is a process in which temperature is constant ,heat can be transferred but temperature is constant and as the temperature is constant so internal energy is also constant
In other hand in adiabatic process there no transfer of heat and internal energy also changes
So the given statement is false statement
Answers
Answer:
I want to believe the program is to be written in java and i hope your question is complete. The code is in the explanation section below
Explanation:
import java.util.Date;
public interface Downloadable {
//abstract methods
public String getUrl();
public Date getLastDownloadDate();
}
Answers
Answer:
It is important to add presentation layer after the physical layer, so that the data along with it's headers can be translated, when the receiver machine is applying a set of different characters.
Data compression is also required to reduce the space that is occupied by data during transmission, now once the presentation is added to the physical layer, data from the physical layer can be compressed at the presentation layer and sent by improving the throughput.
Explanation:
Solution
The presentation of data involves the following as shown below:
Presentation of data comprises of the task like translating between receiver and sender devices so that machines with different capabilities sets can communicate with one another.
It involves encoding and decoding of data to provide data security that is been transmitted by different machines.
Data sometimes needs to compressed for efficiency improvement for transmission.
The physical layer of the TCP/IP protocol suite is responsible or refers to the transmission of physical data over a physical medium
It is good or important to add presentation layer after the physical layer, so that the data along with it's headers can be translated, when the receiver machine is applying a set of different characters.
Data encryption at this stage is good for security instead of encrypting the data at upper/higher layers.
Hence, it is advisable to add presentation layer after the physical layer in the TCP/IP suite.
Answer:
The layer ought to be embedded between Layer 2 and 3.
Explanation:
Applications often communicate with each other. This cannot be successful if they don't see data the same way. The Presentation Layer in the Open Systems Interconnection defines how data is presented and is often processed in the TCP/IP applications.
While the Presentation Layer does not exist as a different layer in the TCP/IP protocol order of arrangement, it is important to note that the Network Layer is also known referred to as the TCP/IP’s Network Layer.
Therefore, if the presentation of the data layer will be separated, it should be between layer 2 and 3.
Cheers!
Answers
Answer:
the magnitude of F_A is 752 N
the direction theta of F_A is 57.9°
Explanations:
Given that,
Resultant force = 1330 N in x direction
∑Fx = R
from the diagram of the question which i uploaded along with this answer
FB = 800 N
FAsin∅ + FBcos30 = 1330 N
FAsin∅ = 1330 - (800 × cos30)
FA = 637.18 / sin∅
Now ∑Fx = 0
FAcos∅ - FBsin30 = 0
we substitute for FA
(637.18 / sin∅)cos∅ = 800 × sin30
637.18 / 800 × sin30 = sin∅/cos∅
and we know that { sin∅/cos∅ = tan∅)
so tan∅ = 1.59295
∅ = 57.88° ≈ 57.9°
THEREFORE FROM THE EQUATION
FA = 637.18 / sin∅
we substitute ∅
so FA = 637.18 / sin57.88
FA = 752 N
Answers
Answer: c) 450 kPa
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
where,
= initial pressure of gas = 150 kPa
= final pressure of gas = ?
= initial volume of gas = v L
= final volume of gas =
Therefore, the new pressure of the gas will be 450 kPa.
Answers
Answer:
mV
Explanation:
The voltage across a capacitor at a time t, is given by:
----------------(i)
Where;
v(t) = voltage at time t
= initial time
C = capacitance of the capacitor
i(t) = current through the capacitor at time t
v(t₀) = voltage at initial time.
From the question:
C = 2μF = 2 x 10⁻⁶F
i(t) = 3 mA
t₀ = 0
v(t₀ = 0) = 0
Substitute these values into equation (i) as follows;
[Solve the integral]
Therefore, the voltage across the capacitor is mV