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Otrada [13]
1 year ago
8

A bucket of cement masse 40kg is tied to the end of the rope connected to a hoist.calculate the tension in the rope when the buc

ket is suspended but stationery
Engineering
1 answer:
Alik [6]1 year ago
4 0

The tension in the rope when this bucket is suspended but stationery (static) is equal to 400 Newton.

<u>Given the following data:</u>

Mass = 40 kg.

<h3>How to calculate the tension?</h3>

Mathematically, the tension in the rope when this bucket is suspended but stationery (static) can be calculated by using this formula:

T = m(a + g)

<u>Note:</u> Acceleration due to gravity (g) is equal to 10 m/s² while the acceleration is 0 m/s² because the bucket is stationery (static)

Substituting the parameters into the formula, we have;

T = 40(0 + 10)

T = 400 Newton.

Read more on tension here: brainly.com/question/4080400

#SPJ1

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Convert the unit Decimeter (dm) into Micrometer (um).
oksian1 [2.3K]

Answer:

86701 Micrometers.

Explanation:

Multiply 0.86701 dm by 100,000 to get 86701 um.

7 0
3 years ago
A boy weighing 108-lb starts from rest at the bottom A of a 6-percent incline and increases his speed at a constant rate to 7 mi
baherus [9]

Answer:

88.18 W

Explanation:

The weight of the boy is given as 108 lb

Change to kg =108*0.453592= 48.988 kg = 49 kg

The slope is given as 6% , change it to degrees as

6/100 =0.06

tan⁻(0.06)= 3.43°

The boy is travelling at a constant speed up the slope = 7mi/hr

Change 7 mi/h to m/s

7*0.44704 =3.13 m/s

Formula for power P=F*v where

P=power output

F=force

v=velocity

Finding force

F=m*g*sin 3.43°

F=49*9.81*sin 3.43° =28.17

Finding the power out

P=28.17*3.13 =88.18 W

4 0
3 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
2 years ago
What is 4 principles of experimental design
Mrrafil [7]

Answer:

manipulation, control , random assignment, and random selection

Explanation:

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1 year ago
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Liquefied Natural Gas (LNG) is a natural gas in its liquid form that is clear, colorless, odorless, non-corrosive, and non-toxic
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Liquefied Natural Gas (LNG) can be defined as a natural gas which in liquid form appear clear and colorless. It is odorless, non-toxic, and non-corrosive. Therefore, the given statement is A) True.

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  • The combustion of LNG produces carbon dioxide and water vapors.
  • It burns with a least pollution thus called as cleanest fossil fuel.
  • The liquefaction of the natural gas takes place at -160 degree Celsius. The liquefaction of the gas causes it to transport easily in gas tanks.
  • LNG is colorless, and clear.
  • LNG does not possess any smell and it is non-corrosive to metallic tanks.
  • LNG is also non-toxic.

Learn more about natural gas:

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6 0
2 years ago
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