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3 years ago

9

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

N/m2). By what volume has 1.0 m3 of water from the surface of the lake been compressed if it is forced down to this depth

1 answer:

Nezavi [6.7K]3 years ago

4 0

Answer:

the volume that has 1.0m^2 of water from the surface of the lake is 4.3

Explanation:

The computation of the volume is as follows;

The Bulk modulus is

= K = 2.3 × 10^9 Pa

K = - V dP ÷ dV

dV = - V dP ÷ K

dV = - (1 × 101.3 × 1000) ÷ 2.3 × 10^9

= - 4.3 × 10^-3 m^3

So the volume that has 1.0m^2 of water from the surface of the lake is 4.3

Hence the same is relevant and considered

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blank refers to the method of spreading fertilizer evenly over the entire field by hand it is done at the blank stage

Andrej [43]

Answer:

Broadcasting is the method, not sure about the stage it is done in

Explanation:

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4 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

3 years ago

How does the suns energy contribute to the carbon cycle

netineya [11]

Answer:

Plants are a good starting point when looking at the carbon cycle on Earth. Plants have a process called photosynthesis that enables them to take carbon dioxide out of the atmosphere and combine it with water. Using the energy of the Sun, plants make sugars and oxygen molecules.

8 0

3 years ago

Read 2 more answers

vovangra [49]

Answer: 3.5 seconds

EXPLANATION:

Using the formula:
v = u + at
And taking the upwards direction as positive, we have the following information:

u = 35 m/s
a = -10m/s^2 (this is acceleration due to gravity)

At the top of its path, the apple will have a velocity of 0 m/s, therefore:

v = 0m/s

Once you substitute everything into the formula, you get:

0 = 35 + (-10)t

Therefore, t = 35/10 = 3.5 seconds

6 0

3 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

4 years ago