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3 years ago

7

A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu

es to sharpen his axe by holding it against the grindstone until it stops 10 s later What is the average torque exertes by the axe on the grindstone?

1 answer:

kolbaska11 [484]3 years ago

8 0

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

$\omega f = \omega o-\alpha*t$

Solving for α:

$\alpha=\frac{\omega o-\omega f}{t}$

where ωo = 600*2*π/60; ωf = 0; t=10s

$\alpha=6.283rad/s^2$

The sum of torque is:

$\tau=I*\alpha$

$\tau=M*R^2/2*\alpha$

$\tau=0.060 N.m$

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3 years ago

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

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Answer:

$21678.47223\ lbf-in^2$

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Stress is given by

$\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2$

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3 years ago

Read 2 more answers

What does a light bulb marked 100 W mean? (A) Uses 100 A of energy per second. (B) Uses 100 J of current per second. (C) 100 C o

Mandarinka [93]

Answer:

(E) none of these.

Explanation:

first that all we need the meaning of a Watt

<em>" is a unit of power. In the International System of Units (SI) it is defined as a derived unit of 1 </em><em>joule</em><em> per </em><em>second</em><em>,[1] and is used to quantify the rate of </em><em>energy</em><em> transfer." </em>

<em>definition taken from wikipedia.org</em>

now with this concept we have that:

100 W means 100 J/s .Therefore it uses 100 J energy per second

therefore on the answers don't mention this three terms together

Joules, energy and second so the answer must be.

(E) None of these

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4 years ago

A heat exchanger is used to heat cold water at 8 C entering at a rate of 1.2 kg/s by hot air at 90 C entering at rate of 2.5 kg/

aleksklad [387]

Answer:

The exit temperature of cold water is 37.6°C

Explanation:

Mass flow rate of water, $\dot{m}_{water} = 1.2 kg/s$

Mass flow rate of air, $\dot{m}_{air} = 2.5 kg/s$

$T_{in} = 90^{0}C\\ T_{out} = 20^0C$

$c_{air} = 1010\\c_{water} = 4186$

$\dot{Q}_{lost} = 28000 J/s$

Applying the conservation of heat law between air, water and the surrounding:

$[\dot{m}c(T_{in} - T_{out})]_{air} = [\dot{m}c(T_{in} - T_{out})]_{water} + \dot{Q}_{lost}$

We get the exit temperature of cold water by making $T_{out}$ the subject of the formula above:

$(T_{out})_{water} =\frac{ [\dot{m}c(T_{in} - T_{out})]_{air} + [\dot{m} c T_{in}]_{water} - \dot{Q}_{lost}}{(\dot{m}c)_{water}} \\\\(T_{out})_{water} = \frac{(2.5*1010*(90-20) + [1.2*4186*(8+273)] - 28000}{1.2*4186} \\\\(T_{out})_{water} = 310.6 K\\\\(T_{out})_{water} = 37.6^0 C$

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3 years ago