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4 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]4 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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The expression with an AND operator ________ when the expression on its left is false and so it does not evaluate the expression

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4 years ago

What do you own that might not be manufactured?

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3 years ago

Steel bar A of 5 mm diameter is subject to a stress of 500 MPa. Aluminum bar B of 10 mm diameter is subject to a stress of 150 M

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Answer:

aluminum bar carrying a higher load than steel bar

Explanation:

Given data;

steel abr

diameter = 5 mm

stress = 500 MPa

aluminium bar

diameter = 10 mm

stress = 150 MPa

we know

stress = laod/area

for steel bar

$500 = \frac{P}{\frac{\pi}{4} 5^2}$

solving for P

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aluminum bar carrying a higher load than steel bar

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3 years ago

Air,in a piston cylinder assembly, is initially at 300 K and 200 kPa.It is then heated at constant pressure to 600 K. Determine

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5 0

3 years ago

Read 2 more answers

Calculate the De-Broglie wavelength for electron accelerated across a potential of 10V, 100V, 1000V, and 10kV respectively. Whic

Arlecino [84]

Answer:

a) $\lambda=3.87 \r A$

b) $\lambda=1.23 \r A$

c) $\lambda=0.387 \r A$

d) $\lambda=0.123 \r A$

Explanation:

We know that the kinetic energy can be expressed in terms of momentum (p= mv):

$K=\frac{p^{2}}{2m}$

p is the momentum
m is the mass of the particle

So, the momentum will be:

$p=\sqrt{2mK}$ (1)

We can use the energy conservation to relate K and the electric potential energy. We assume that all potential energy becomes kinetic energy. Therefore:

$K=q_{e}V$ (2)

a) If V=10 [V], K will be:

$K=1.6*10^{-19}*10=1.6*10^{-18}[J]$

We can find the momentum using the equation (1)

$p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-18}}=1.71*10^{-24}[kg*m*s^{-1}]$

The De-Broglie wavelength equation is given by:

$\lambda=\frac{h}{p}$ (3)

h is the Plank constant ( $h=6.626 x 10^{-34} [J*s]$ )

$\lambda=\frac{6.626 x 10^{-34}}{1.71*10^{-24}}=3.87*10^{-10}[m]=3.87 \r A$

b) If V=100 [V], using the same analyze, the De-Broglie wavelength will be:

$K=1.6*10^{-19}*100=1.6*10^{-17}[J]$

$p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-17}}=5.39*10^{-24}[kg*m*s^{-1}]$

$\lambda=\frac{6.626 x 10^{-34}}{5.39*10^{-24}}=1.23*10^{-10}[m]=1.23 \r A$

c) V=1000 [V]

$K=1.6*10^{-19}*100=1.6*10^{-16}[J]$

$p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-16}}=1.71*10^{-23}[kg*m*s^{-1}]$

$\lambda=\frac{6.626 x 10^{-34}}{1.71*10^{-23}}=3.87*10^{-11}[m]=0.387 \r A$

d) V=10000 [V]

$K=1.6*10^{-19}*100=1.6*10^{-15}[J]$

$p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-15}}=5.40*10^{-23}[kg*m*s^{-1}]$

$\lambda=\frac{6.626 x 10^{-34}}{5.40*10^{-23}}=1.23*10^{-11}[m]=0.123 \r A$

The fringe spacing interference is proportional to the wavelength, so in our case, the larger fringe spacing occurs when voltage is 10 V, here λ = 3.87 angstroms.

I hope it helps you!

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3 years ago