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4 years ago

When an object is moving, we use the following coefficient for friction calculations a)-μk b)-μs c)-γk d)- γs

Engineering

1 answer:

Reika [66]4 years ago

3 0

Answer: $\mu_{k}$

Explanation:

We use kinetic friction when a body is moving i.e. $\mu_{k}$ for calculations.

Static friction is used when a body is in rest while kinetic friction is used when a body is moving and its value is quite low as compared to static friction .

Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.

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Why is the face of the claw on a claw hammer usually a smooth curve? Why isn't it straight or some other shape?

GarryVolchara [31]

Answer:

The face of the claw on the claw hammer is usually a smooth curve so as to improve the ease with which nails are removed when removing nails because as the nail held between the V shaped split claw is being pulled out from the wood, it slides more and more towards cheek, reducing the distance of the nail from the cheek which is the fulcrum, thereby increasing the mechanical advantage because the location of the hand on the grip remains unchanged

Explanation:

7 0

3 years ago

The volume of 1.5 kg of helium in a frictionless piston-cylinder device is initially 6 m3. Now, helium is compressed to 2 m3 whi

coldgirl [10]

Answer:

The initial temperature will be "385.1°K" as well as final will be "128.3°K".

Explanation:

The given values are:

Helium's initial volume, v₁ = 6 m³

Mass, m = 1.5 kg

Final volume, v₂ = 2 m³

Pressure, P = 200 kPa

As we know,

Work, $W=p(v_{2}-v_{1})$

On putting the estimated values, we get

⇒ $=200000(2-6)$

⇒ $=200000\times (-4)$

⇒ $=800,000 \ N.m$

Now,

Gas ideal equation will be:

⇒ $pv_{1}=mRT_{1}$

On putting the values. we get

⇒ $200000\times 6=1.5\times 2077\times T_{1}$

⇒ $T_{1}=\frac{1200000}{3115.5}$

⇒ $=385.1^{\circ}K$ (Initial temperature of helium)

and,

⇒ $pv_{2}=mRT_{2}$

On putting the values, we get

⇒ $200000\times 2=1.5\times 2077\times T_{2}$

⇒ $T_{2}=\frac{400000}{3115.5}$

⇒ $=128.3^{\circ}K$ (Final temperature of helium)

3 0

3 years ago

Since the passing of the Utah GDL laws in 1999:

wlad13 [49]

The answer is b, I hope this helps you

7 0

3 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$