Which explanation best summarizes what went wrong during Paul’s cost analysis?

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nikklg [1K]

Answer:

I did not what is it about?

8 0

2 years ago

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The driveshaft of an automobile is being designed to transmit 134 hp at 2900 rpm. Determine the minimum diameter d required for

Misha Larkins [42]

Answer:

The minimum diameter is 1.344 in

Explanation:

The angular speed of the driveshaft is equal to:

$w=\frac{2\pi N}{60}$

Where

N = rotational speed of the driveshaft = 2900 rpm

$w=\frac{2\pi *2900}{60} =303.69rad/s$

The torque in the driveshaft is equal to:

$\tau=\frac{P}{w}$

Where

P = power transmitted by the driveshaft = 134 hp = 73700 lb*ft/s

$\tau=\frac{73700}{303.69} =242.68lb*ft$

The minimum diameter is equal to:

$d_{min} =(\frac{16T}{\pi *\tau } )^{1/3}$

Where

T = shear stress = 6100 psi

τ = 242.68 lb*ft = 2912.16 lb*in

$d_{min} =(\frac{16*2912.16}{\pi *6100} )^{1/3} =1.344in$

4 0

2 years ago

The velocity of a particle which moves along the s-axis is given by v = 2-4t+5t^(3/2), where t is in seconds and v is in meters

Brut [27]

Answer:

a) s = 42 m

b) V = 26 m/s

c) a = 11 m/s²

Explanation:

The velocity(V) = 2 - 4t + 5t^3/2 where t is in seconds and V is in m/s

a) To get the position, we have to integrate the velocity with respect to time.

We get the position(s) from the equation:

$V=\frac{ds}{dt}$

$ds=Vdt$

Integrating both sides,

$s=\int\ {V} \, dt$

$s=\int\ {(2-4t+5t^{\frac{3}{2} }) \, dt$

Integrating, we get

$s=(2t-2t^{2} +2t^{\frac{5}{2} }+c)m$

But at t=0, s(0) = 2 m

Therefore,

$s(0)=2(0)-2(0)^{2} +2(0)^{\frac{5}{2} }+c$

$2=0+0+0+c$

c = 2

Therefore

$s=(2t-2t^{2} +\frac{10}{5}t^{\frac{5}{2} }+2)m$

When t = 4,

$s=2(4)-2(4)^{2} +2(4)^{\frac{5}{2} }+2$

$s=8-32+64+2$

s = 42 m

At t= 4s, the position s = 42m

b) The velocity equation is given by:

$V=(2-4t+5t^{\frac{3}{2} })m/s$

At t = 4,

$V=2-4(4)+5(4)^{\frac{3}{2} }$

V = 2 - 16 + 40 = 26 m/s

The velocity(V)at t = 4 s is 26 m/s

c) The acceleration(a) is given by:

$a=\frac{dv}{dt}$

$a=\frac{d}{dt}(2-4t+5t^{ \frac{3}{2}})$

Differentiating with respect to t,

$a=-4+\frac{15}{2}t^{\frac{1}{2} }$

$a=(-4+\frac{15}{2}t^{\frac{1}{2} })m/s^{2}$

At t = 4 s,

$a=-4+\frac{15}{2}(4)^{\frac{1}{2} }$

$a=-4+15$

a = 11 m/s²

4 0

2 years ago

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SPELL HELLO PLZ I NEED HELP BIG TIME

QveST [7]

Answer:

hello

Explanation:

8 0

2 years ago

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What type of power source is used for gmaw

nataly862011 [7]

Answer:

The GMAW process commonly uses a constant voltage power source (GMAW-CV) that allows for a relatively constant welding voltage output over a range of welding currents. For GMAW-CV the welder selects the wire feed speed (WFS) on the wire feeder unit and an appropriate voltage on the welding power supply.

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2 years ago

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