True or false? Don't break or crush mercury-containing lamps because mercury powder may be released.

A 4-m-high and 6-m-long wall is constructed of twolarge 2-cm-thick steel plates (k 5 15 W/m·K) separated by1-cm-thick and 20-cm

Flura [38]

Answer:

fart

Explanation:

6 0

3 years ago

Water exerts little pressure on a building so it has no implications on foundation design.

777dan777 [17]

Fact

Explanation:

If wrong Im sorry Correct me In The comment please

7 0

3 years ago

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

natima [27]

Answer:

1) A=282.6 mm

2) $a_{max}=60.35\ m/s^2$

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm ------------1

V=2.5 m/s at x=225 mm ------------2

Where x measured from mid position.

We know that velocity in simple harmonic given as

$V=\omega \sqrt{A^2-x^2}$

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

$3.5=\omega \sqrt{A^2-0.15^2}$ ------3

$2.5=\omega \sqrt{A^2-0.225^2}$ --------4

Now by dividing equation 3 by 4

$\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}$

$1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}$

So A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

$3.5=\omega \sqrt{A^2-0.15^2}$

$3.5=\omega \sqrt{0.2826^2-0.15^2}$

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

$a_{max}=\omega ^2A$

$a_{max}=14.61 ^2\times 0.2826\ m/s^2$

$a_{max}=60.35\ m/s^2$

Time period T

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{14.609}$

T=0.42 sec

8 0

3 years ago

..........................

kow [346]

Answer:

i wish i knew what you meant by that

7 0

3 years ago

10.0 kmol of a 40.0 mol% methanol and 60.0 mol% water mixture is processed in a normal batch distillation system with a still po

Serggg [28]

Answer:

[a]. 0.49.

[2]. 0.536

[c]. 4.15 kmol; 5.84 kmol.

Explanation:

Without mincing words let's dive straight into the solution to the question above.

[a].

The initial external reflux ratio, LD that must be used = [(0.85 - 0.57)/ 0.85 - 0]/ [ 1 - (0.85 - 0.57)/ 0.85 - 0].

The initial external reflux ratio, LD that must be used = 0.329/ 1- 0.329 = 0.49.

[b].

The final external reflux ratio that must be used = [ 0.85 - 0.13/ 0.85 - 0]/ [ 1 - 0.85 - 0.13/ 0.85 - 0].

Hence, the final external reflux ratio that must be used =0.847/ 1 - 0.847 = 5.536.

[c].

The amount of distillate product that is withdrawn:

4 = 0.85 H(t) + 0.8 - 0.08.

H(t) = 4.15 kmol, and the value of Wfinal = 5.84 kmol.

3 0

3 years ago