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MrMuchimi
3 years ago
12

The force on a cutting tool are 2600N vertically downward and 2100 horizontal. Determine the resultant force acting on the tool

and the angle at which it acts.​
Engineering
1 answer:
kolezko [41]3 years ago
3 0

Answer & Explanation:

"The force <em>on a cutting tool</em> are 2600N vertically downward" sounds a little unusual, since most of the time, the tool is above the object to be cut in such a way that the force acting "ON the tool" is upwards.  We will accept the statement as it is (downwards).

Since the two forces are acting at right angles to each other, the resultant can be found using Pythagoras theorem, namely

resultant = sqrt(2600^2+2100^2) = 3342 N (approx.)

The angle can be found using the arcTangent function, or

angle = arcTangent(2600/2100) = 51.07 degrees below the horizontal, since the 2600 N force is acting downwards.

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sineoko [7]

Answer:

The correct option is;

c.  the exergy of the tank can be anything between zero to P₀·V

Explanation:

The given parameters are;

The volume of the tank = V

The pressure in the tank = 0 Pascal

The pressure of the surrounding = P₀

The temperature of the surrounding = T₀

Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change

The exergy balance equation  is given as follows;

X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}

Where;

X₂ - X₁ is the difference between the two exergies

Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system

Therefore, the exergy of the tank can be anything between zero to P₀·V.

6 0
3 years ago
determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004
Afina-wow [57]

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 ,  To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

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