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antoniya [11.8K]
3 years ago
15

Here some free points

Engineering
2 answers:
Vanyuwa [196]3 years ago
6 0

Answer:

I will accept your friend request I hope you also follow me back

Semenov [28]3 years ago
5 0
Oh hehehhe yayayay yas in ssnsjd
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Early mountaineers boiled water to estimate their altitude. If they reached the top and found that water boiled at 89°C, approxi
algol [13]

Answer:

The height of the mountain is 4670.35 metres.

Explanation:

8 0
4 years ago
If a motorist moves with a speed of 30 km/hr, and covers the distance from place A to place B
Sergio039 [100]

Answer:

105 km

Explanation:

The motorist was going 30 km/hr, and it took 3 hours 30 minutes. That's 3.5 hours. 3.5×30=105

5 0
4 years ago
Question in Probability and Statistics 2<br>​
Ghella [55]
Since we both have a question in same subject may u help me with this For a confidence level of 80% with a sample size of 21, find the critical t value
7 0
3 years ago
Consider an open loop 1-degree-of-freedom mass-spring damper system. The system has mass 4.2 kg, and spring stiffness of 85.9 N/
Marat540 [252]

Answer:

Damping ratio  \zeta =0.0342

Explanation:

Given that

m=4.2 kg,K=85.9 N/m,C=1.3 N.s/m

We need to find damping ratio

We know that critical damping co-efficient

 C_c=2\sqrt {mk}

 C_c=2\sqrt {4.2\times 85.9}

 C_c=37.98 N.s/m

Damping ratio(\zeta) is the ratio of damping co-efficient to the critical damping co-efficient

So \zeta =\dfrac{C}{C_c}

\zeta =\dfrac{1.3}{37.98}

\zeta =0.0342

So damping ratio  \zeta =0.0342

 

3 0
4 years ago
A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L
zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

3 0
4 years ago
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