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3 years ago

15

Which of the following natural processes is most likely to support the formation of an underwater sinkhole

1 answer:

ExtremeBDS [4]3 years ago

3 0

Pollution build up made from deposited minerals , limestone cave collapsing because the sea levels changed ,

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Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.

nikitadnepr [17]

Answer:

a) $m_e= 3.05 Kg$

b) $\rho=1072.3kg/m^3$

c) $m_e= 3.05 Kg$

Explanation:

From the question we are told that:

Beaker Mass $m_b=1.20$

Liquid Mass $m_l=1.85$

Balance D:

Mass $m_d=3.10$

Balance E:

Mass $m_e=7.50$

Volume $v=4.15*10^{-3}m^3$

a)

Generally the equation for Liquid's density is mathematically given by

$m_e=m_b+m_l+(\rho*v)$

$\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}$

$\rho=1072.3kg/m^3$

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

$m_d=m- (\rho *v )$

$m=3.10+ (1072.30 *4.15*10^{-3}m^3 )$

$m=18.10kg$

c)

Generally the equation for E's Reading at A pulled is mathematically given by

$m_e=m_b+m_l$

$m_e = 1.20 + 1.85$

$m_e= 3.05 Kg$

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3 years ago

Advantages, disadvantages and applications of dsb-sc

allochka39001 [22]

Answer:

<h3>advantages: </h3>

<em>lower power consumption, modulation system is simple</em>

<h3>disadvantages<em>:</em></h3>

<em>complex detection</em>

<h3><em>applications:</em></h3>

analog TV systems: to transmit color information

<h3><em /></h3>

<em />

<em />

<em />

<em />

Explanation:

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4 years ago

Modify the sentence-generator program so that it inputs its vocabulary from a set of text files at startup. The filenames are no

OlgaM077 [116]

Answer:

1

Explanation:

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2 years ago

Please help me, I need help.

elixir [45]

Answer:

umm

Explanation:

nothing post so how we post to help you

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3 years ago

Consider a Carnot cycle executed in a closed system with 0.0058 kg of air. The temperature limits of the cycle are300 K and 940

cricket20 [7]

Answer:0.646 KJ

Explanation:

Using First law for cycle

$\sum Q=\sum W$

$\sum Q=Q_{1-2}+Q_{3-4}$

For adiabatic process heat transfer is zero and for isothermal process

d(Q)=d(W)

$Q_{1-2}=mRT_1\ln {\frac{P_1}{P_2}}$

Given $P_1=2000KPa$

$P_3=20KPa$

$\left (\frac{T_2}{T_3}\right )^{\frac{\gamma }{\gamma -1}$ = $\left (\frac{P_2}{P_3}\right )}$

$P_2=1089.06K$

$Q_{1-2}=0.0058\dot 0.287\dot 940\ln \frac{2000}{1089.06}$ = $0.95KJ$

$Q_{3-4}=mRT_2\ln {\frac{P_3}{P_4}}$

$\left (\frac{T_1}{T_4}\right )^{\frac{\gamma }{\gamma -1}$ = $\left (\frac{P_1}{P_4}\right )}$

Now we have to find $P_4=36.72KPa$

$Q_{3-4}=0.0058\dot 0.287\dot 300\ln \frac{20}{36.72}$ = $-0.30341KJ$

$Q_{net}=Q_{1-2}+Q_{3-4}$

$Q_{net}=0.95-0.303=0.646KJ$

$Q_{net}=W_{net}=0.646KJ$

7 0

3 years ago