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2 years ago

How is engine power expressed?

2 answers:

6 0

Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.

mars1129 [50]2 years ago

5 0

Answer: Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.

Explanation: (I hope this helped!! ^^)

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The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc

Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.

Given Data

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist

∅=4°

∅=4* $\pi$ /180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist

∅=TL/GJ

∅=TL/G* $\pi$ /2*c^4

∅=2TL/G* $\pi$ *c^4

c= $\sqrt[4]{2TL/\pi G }$ ∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J* $\pi$ /2*c^4)]*c

г =[2T/(J* $\pi$ *c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required

We will use larger of the two values

c= 18.0569 x 10^-3 m

c = 18.0569 mm

3 0

3 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

3 years ago

How to code the round maze in CoderZ?

dlinn [17]

Answer:

hola

Explanation:

5 0

2 years ago

What is the mechanical advantage of a pulley with 3 support ropes?

snow_tiger [21]

Answer:

The mechanical advantage is 3 to 1

Explanation:

A frictionless pulley with three support ropes carries equal tension on each of the ropes thus;

Tension in each pulley rope = T

Total tension in the 3 ropes = 3 × T = 3·T

Direction of the tension forces on each rope = Unidirectional

Total force provided by the 3 ropes = 3·T

Therefore, a force, T, applied at the end of the rope will result in a lifting force of 3·T

Hence, the mechanical advantage = 3·T to T which is presented as follows;

$Mechanical \ advantage = \dfrac{3 \cdot T}{T} = \dfrac{3}{1}$

The mechanical advantage = 3 to 1.

5 0

3 years ago

How does a catapult incorporate technology?

storchak [24]

I’m sure the answer is correct and it is below
“ A catapult works because energy can be converted from one type to another and transferred from one object to another. ... This energy is stored in the launching device as potential, or stored, energy. The catapult you are about to make uses elastic potential energy stored in a wooden stick as you bend it.”

6 0

3 years ago