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3 years ago

Torque is a twisting force. If the required torque applied on a 3 ft wrench is 45 ft·lb, what is the force that must be applied?

Engineering

1 answer:

natka813 [3]3 years ago

8 0

Answer:

15 lbs

Explanation:

assuming you push from the end of the wrench (3ft)

torque = force(distance)

force = torque/distance

(45 ft·lb)/(3 ft)= 15 lbs

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A heat engine receives heat from a heat source at 1453 C and has a thermal efficiency of 43 percent. The heat engine does maximu

xxMikexx [17]

Answer:

a) 1253 kJ

b) 714 kJ

c) 946 C

Explanation:

The thermal efficiency is given by this equation

η = L/Q1

Where

η: thermal efficiency

L: useful work

Q1: heat taken from the heat source

Rearranging:

Q1 = L/η

Replacing

Q1 = 539 / 0.43 = 1253 kJ

The first law of thermodynamics states that:

Q = L + ΔU

For a machine working in cycles ΔU is zero between homologous parts of the cycle.

Also we must remember that we count heat entering the system as positiv and heat leaving as negative.

We split the heat on the part that enters and the part that leaves.

Q1 + Q2 = L + 0

Q2 = L - Q1

Q2 = 539 - 1253 = -714 kJ

TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:

η = 1 - T2/T1

T2/T1 = 1 - η

T2 = (1 - η) * T1

The temperatures must be given in absolute scale (1453 C = 1180 K)

T2 = (1 - 0.43) * 1180 = 673 K

673 K = 946 C

8 0

3 years ago

For each of the resistors shown below, use Ohm's law to calculate the unknown quantity, Be sure to put your answer in proper eng

daser333 [38]

Answer:

the hurts my brain sorry bud cant help

Explanation:

6 0

3 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .