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4 years ago

Why doesn't glue stick to the inside of the bottle?

Engineering

1 answer:

Galina-37 [17]4 years ago

6 0

Answer:

Explanation:

When white is inside a , there's not enough air inside the to cause the water to

to make the sticky. Basically, the protects the from the air and keeps the runny.

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You have designed a treatment system for contaminant Z. The treatment system consists of a pipe that feeds into a CSTR. The pipe

IRISSAK [1]

Answer:

0.667 per day.

Explanation:

Our values here are

$Q=10m^3/dV_p=15m^3\\V_{cstr}=60m^3\\c_p=2500mg/L\\c_{cstr}=500mg/L$

Degradation constant=k and is unknown.

We calculate the concentration through the formula,

$cc_{cstr} =\frac{c_{in}}{1+K(V/Q)} \\cc_{cstr}=\frac{c_p}{1+K*\frac{V_{csrt}}{Q}}$

Replacing values we have

$1+k(\frac{60}{10})=\frac{2500}{500}\\1+k=5\\K(6)=5-1\\K(6)=4\\K=2/3\\K=0.667/day$

That is the degradation constant of Z-contaminant

3 0

3 years ago

A computer has a two-level cache. Suppose that 60% of the memory references hit on the first level cache, 35% hit on the second

Andreyy89

Answer:

t=14ns

Explanation:

We make the relation between the specific access time and the memory percentage in each level, so

$60\% \Rightarrow 60/100 = 0.60\\35\% \Rightarrow 35/100 = 0.35\\05\% \Rightarrow 05/100 = 0.05$

$t= 0.6(5) + 0.35(5+15) + 0.05(5+15+60)\\t= 0.6(5) + 0.35(20) + 0.05(80)\\t= 3 + 7 + 4\\t= 14 ns$

Average Access Time is 14 nsec.

4 0

4 years ago

3. (9 points) A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 10

Bezzdna [24]

Answer:

a) 49.95 watts

b) The self locking condition is satisfied

Explanation:

Given data

weight of the square-thread power screw ( w ) = 100 kg = 1000 N

diameter (d) = 20 mm ,

pitch (p) = 2 mm

friction coefficient of steel parts ( f ) = 0.1

Gravity constant ( g ) = 10 N/kg

Rotation of electric power screwdrivers = 300 rpm

A ) Determine the power needed to raise to the basket board

first we have to calculate T

T = Wtan (∝ + Ф ) * $\frac{Dm}{2}$ ------------- equation 1

Dm = d - 0.5 ( 2) = 19mm

Tan ∝ = $\frac{L}{\pi Dm}$ where L = 2*2 = 4

hence ∝ = 3.83⁰

given f = 0.1 , Tan Ф = 0.1. hence Ф = 5.71⁰

insert all the values into equation 1

T = 1.59 Nm

Determine the power needed using this equation

$\frac{2\pi NT }{60}$ = $\frac{2\pi * 300 * 1.59}{60}$

= 49.95 watts

B) checking if the self-locking condition of the power screw is satisfied

Ф > ∝ hence it is self locking condition is satisfied

3 0

3 years ago

A water skier leaves the end of an 8 foot tall ski ramp with a speed of 20 mi/hr and at an angle of 250. He lets go of the tow r

klemol [59]

Answer:

At highest point:

y1 = 10.4 ft

v1 = (26.5*i + 0*j) ft/s

When he lands:

x2 = 31.5 ft (distance he travels)

t2 = 1.19 s

V2 = (26.5*i - 25.9*j) ft/s

a2 = -44.3°

Explanation:

Since he let go of the tow rope upon leaving the ramp he is in free fall from that moment on. In free fall he is affected only by the acceleration of gravity. Gravity has a vertical component only, so the movement will be at constant acceleration in the vertical component and at constant speed in the horizontal component.

20 mi / h = 29.3 ft/s

If the ramp has an angle of 25 degrees, the speed is

v0 = (29.3 * cos(25) * i + 29.3 * sin(25) * j) ft/s

v0 = (26.5*i + 12.4*j) ft/s

I set up the coordinate system with the origin at the base of the ramp under its end, so:

R0 = (0*i + 8*j) ft

The equation for the horizontal position is:

X(t) = X0 + Vx0 * t

The equation for horizontal speed is:

Vx(t) = Vx0

The equation for vertical position is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

The equation for vertical speed is:

Vy(t) = Vy0 + a * t

In this frame of reference a is the acceleration of gravity and its values is -32.2 ft/s^2.

In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.

0 = Vy0 + a * t1

a * t1 = -Vy0

t1 = -Vy0 / a

t1 = -12.4 / -32.2 = 0.38 s

y1 = y(0.38) = 8 + 12.4 * 0.38 + 1/2 * (-32.2) * (0.38)^2 = 10.4 ft

The velocity at that moment will be:

v1 = (26.5*i + 0*j) ft/s

When he lands in the water his height is zero.

0 = 8 + 12.4 * t2 + 1/2 * (-32.2) * t2^2

-16.1 * t2^2 + 12.4 * t2 + 8 = 0

Solving this equation electronically:

t2 = 1.19 s

Replacing this time on the position equation:

X(1.19) = 26.5 * 1.19 = 31.5 ft

The speed is:

Vx2 = 26.5 ft/s

Vy2 = 12.4 - 32.2 * 1.19 = -25.9 ft/s

V2 = (26.5*i - 25.9*j) ft/s

a2 = arctg(-25.9 / 26.5) = -44.3

3 0

3 years ago

g (d) Usually, in the case of two finite-duration signals like in parts (a) and (b), the convolution integralmust be evaluated i

Bas_tet [7]

you face is A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

3 0

4 years ago