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Mekhanik [1.2K]

3 years ago

8

Solar evaporation ponds are shallow so that the sun can heat them up, causing the water to evaporate, which increases the salt c

oncentration. A hybrid of which two bacteria tested in this lab would find life in this habitat ideal?

1 answer:

ehidna [41]3 years ago

6 0

Answer:

B. stearothermophilus and S. ruber

Explanation:

B. stearothermophilus and S. ruber

In solar evaporation ponds the temperature is higher and the salt concentration is also higher because of the water evaporated so sunder such extreme conditions this hybrid bacteria is capable of surviving. B. stearothermophilus is thermophilus bacteria which grows at high temperature and S. ruber is halophilic bacteria which grows in saline environment. So, these two bacteria best suited for the above hybrid condition.

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1.2 Define the following terms and in each case give the symbol and the unit: 1.2.1 wavelength (4) ·

Alex777 [14]

Wavelength is represented by lambda( $\lambda$ )

$\\ \bull\tt\longmapsto \lambda=\dfrac{c}{v}$

C is speed of light in air.
v is the frequency.

It has unit as meter(m)

8 0

3 years ago

The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?

postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

$Initial\ Distance = 2cm$ ---- $r_1$

$New\ Distance = 4cm$ --- $r_2$

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

$F_1 = \frac{kq_1q_2}{r_1^2}$ --- Coulomb's law

Substitute 2 for r1

$F_1 = \frac{kq_1q_2}{2^2}$

$F_1 = \frac{kq_1q_2}{4}$

The new force (F2) is

$F_2 = \frac{kq_1q_2}{r_2^2}$

Substitute 4 for r2

$F_2 = \frac{kq_1q_2}{4^2}$

$F_2 = \frac{kq_1q_2}{4*4}$

$F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}$

Substitute $F_1 = \frac{kq_1q_2}{4}$

$F_2 = \frac{1}{4}*F_1$

$F_2 = \frac{F_1}{4}$

The new force is 1/4 of the previous force.

3 0

3 years ago

Which stretch focuses on stretching the glutes and hamstrings while laying on your back?

WINSTONCH [101]

Answer:

knee to chest

Explanation:

I took the test

5 0

3 years ago

Read 2 more answers

The weightlifter's internal store of energy decreased when he lifted the bar.

Mumz [18]

Answer:

The energy returns to the weightlifter's muscles, where it is dissipated as heat.

Explanation:

The energy returns to the weightlifter's muscles, where it is dissipated as heat. As long as the weightlifter controls the weight's descent, their muscles are acting as an overdamped shock absorber, as if the weight were sitting on a piston containing very thick fluid, slowly compressing it downward (and slightly heating up the fluid in the process). Since muscles are complicated biological systems and not simple pistons, they require metabolic energy to maintain tension throughout the controlled descent, so the weightlifter feels like they're putting energy into the weight, even though the weight's gravitational potential energy is being converted into heat within the lifter's muscles.

5 0

3 years ago

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

5 0

3 years ago