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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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What's the price of MDF wood? Is it cheap or expensive?? That's all, thank you.

Mashcka [7]

Answer:

it is gud

Explanation:

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6 0

3 years ago

Calculate the time taken to completely empty aswimming pool 15

user100 [1]

Answer:

Time needed to empty the pool is 401.35 seconds.

Explanation:

The exit velocity of the water from the orifice is obtained from the Torricelli's law as

$V_{exit}=\sqrt{2gh}$

where

'h' is the head under which the flow of water occurs

Thus the theoretical discharge through the orifice equals

$Q_{th}=A_{orifice}\times \sqrt{2gh}$

Now we know that

$C_{d}=\frac{Q_{act}}{Q_{th}}$

Thus using this relation we obtain

$Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}$

Now we know by definition of discharge

$Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}$

Using the above relations we obtain

$Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\$

The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0

Solving for 't' we get

$\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.$

4 0

3 years ago

Sin x +√3 coax= √2

Ber [7]

Answer:

The two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12

Explanation:

The given equation is

Sin x +√3 Cosx= √2

Upon dividing the equation by 2 we get

$\frac{1}{2}Sinx + \frac{\sqrt{3} }{2}Cosx = \frac{\sqrt{2} }{2}$

Sin( $\frac{pi}{6}$ )*Sinx + Cos( $\frac{pi}{6}$ )*Cosx = $\frac{1}{\sqrt{2} }$

This makes the formula of

CosACosB + SinASinB = Cos(A-B)

Cos(x- $\frac{pi}{6}$ ) = $\frac{1}{\sqrt{2} }$

cos(x- pi/6) = cos(pi/4)

upon writing the general equation we get

x-pi/6 = 2n*pi ± pi/4

x = 2n*pi ± pi/4 -pi/6

so we will have two solutions

x = 2n*pi + pi/4 -pi/6

= 2n*pi + pi/12

and

x = 2n*pi - pi/4 -pi/6

= 2n*pi -5pi/12

Therefore the two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12.

4 0

3 years ago

Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of the h

zhenek [66]

Answer:

The Diameter of an atom that fits into the FCC is 2r,

Which leads to the final steps of 0414R

Explanation:

The diameter of an atom that will fit into the FCC site is (2r) which differentiates length edge unit cell for the radii of two atoms that are seen on both side of the site (R); which is 2r = a – 2R

The FCC is also a part of R, therefore a=2R √2 

 Solving for R, the equation is, r =a-2R/2 =2R√2-2R/2 =0414R

4 0

3 years ago

What does snow fall from?

Klio2033 [76]

Answer:

Clouds

Explanation:

It is created by trapped dust and water.

4 0

3 years ago