ENGINEERING COLLEGE

Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Distance between point A and point B is 5.0 meters and they are at the same level. (Ignore the frictional losses)

Answers

Answer 1

Answer:

Answer:

0.556 Watts

Explanation:

w = Weight of object = 762 N

s = Distance = 5 m

t = Time taken = 29 seconds

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$s=ut+(1)/(2)at^2\n\Rightarrow a=(2* (s-ut))/(t^2)\n\Rightarrow a=(2* (5-0))/(29^2)=(10)/(481)$

Mass of the body

$m=(w)/(g)=(762)/(9.81)$

Force required to move the body

$F=ma\n\Righarrow F=(762)/(9.81)* (10)/(481)$

Velocity of object

$v=u+at\n\Rightarrow v=0+(10)/(481)* 29\n\Rightarrow v=(10)/(29)$

Power

$P=Fv\n\Rightarrow P=(762)/(9.81)* (10)/(481)* (10)/(29)=0.556\ W$

∴ Amount of power required to move the object is 0.556 Watts

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"Given a nodal delay of 84.1ms when there is no traffic on the network (i.e. usage = 0%), what is the effective delay when network usage = 39.3% ? (Give answer is miliseconds, rounded to one decimal place, without units. So for an answer of 0.10423 seconds you would enter "104.2" without the qu"

Answers

Answer:

Explanation:

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effective delay = $84.1 * (100)/(100-39.3)=138.55024711697ms$

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operates in a 10°C environment. What is the maximum thermal efficiency? What is it if the collector is redesigned to focus the incoming light to produce a maximum temperature of 300°C?

Answers

Answer:

$\eta _(max) = 0.2413 = 24.13%$

$\eta' _(max) = 0.5061 = 50.61%$

Given:

$T_(1max) = 100^(\circ) = 273 + 100 = 373 K$

operating temperature of heat engine, $T_(2) = 10^(\circ) = 273 + 10 = 283 K$

$T_(3max) = 300^(\circ) = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _(max)$ is given by:

$\eta _(max) = 1 - (T_(2))/(T_(1max))$

$\eta _(max) = 1 - (283)/(373) = 0.24$

$\eta _(max) = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_(3max)$ changes to $300^(\circ)$ , so, the new maximum efficiency, $\eta' _(max)$ is given by:

$\eta' _(max) = 1 - (T_(2))/(T_(3max))$

$\eta _(max) = 1 - (283)/(573) = 0.5061$

$\eta _(max) = 0.5061 = 50.61%$

If you answer the whole question and show your work/coding I will rate 5 stars/brainliest!!! Walnut Orchard has two farms that grow wheat and corn. Because of different soil conditions, there are differences in the yields and costs of growing crops on the two farms. The yields and costs are shown in the following table. Each farm has 100 acres available for cultivation. 11,000 bushels of wheat and 7,000 bushels of corn must be grown. Please have an LP model to minimize the total cost while meeting the demand and solve it with Lindo or Excel. You need to have all parts of a model: notation, objective function, constraints, and sign restrictions.

Answers

Answer: 18,100

Explanation: two farms that grow wheat and corn.

What is the De Broglie wavelength of an electron under 150 V acceleration?

Answers

Answer:

0.1 nm

Explanation:

Potential difference of the electron = 150 V

Mass of electron $m=9.1* 10^(-31)kg$

Charge on electron $1.6* 10^(-19)C$

Plank's constant $h=6.67* 10^(-34)$

If the velocity of the electron is v

Then according to energy conservation $eV =(1)/(2)mv^2$

$v=\sqrt{(2eV)/(m)}=\sqrt{(2* 1.6* 10^(-`19)* 150)/(9.1* 10^(-31))}=7.2627* 10^(6)m/sec$

According to De Broglie $\lambda =(h)/(mv)=(6.67* 10^(-34))/(9.1* 10^(-31)* 7.2627* 10^(6))=0.1nm$

In the hydrodynamic entrance region of a pipe with a steady flow of an incompressible liquid A. The average velocity increases with distance from the entrance.
B. The average velocity stays the same with distance from the entrance.
C. The maximum velocity increases with distance from the entrance.
D. The maximum velocity decreases with distance from the entrance.
E. B and C
F. B and D

Answers

Answer:

D. The maximum velocity decreases with distance from the entrance.

Explanation:

This is because over time, the pressure with with the incompressible liquid enters decreases with distance from the entrance

Answer:

C. The maximum velocity increases with distance from the entrance

Explanation:

As the fluid particles moves into the pipe, the layer of fluid in contact with the surface of the pipe come to a complete stop. This layer also causes the fluid

particles in the adjacent layers to gradually slow down as a result of friction between fluid molecules, leaving the fluid at the center of the pipe with the maximum velocity.

Since the fluid is incompressible, to make up for this velocity reduction, the velocity of the fluid at the mid-

section of the pipe has to increase to keep the mass flow rate through the

pipe constant. As a result, a velocity gradient develops along the pipe and the maximum velocity which is at the center of the pipe increases with distance from entrance.

1. The presentation of data is becoming more and more important in today's Internet. Some people argue that the TCP/IP protocol suite needs to add a new layer to take care of the presentation of data. If this new layer is added in the future, where should its position be in the suite?

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!

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