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3 years ago

HELP ME, PLEASE.

Physics

2 answers:

Wittaler [7]3 years ago

8 0

I would say tempurature

Leno4ka [110]3 years ago

4 0

The temperature determines the star's color.

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Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L − 1 and R max ( V max ) = 222 μmol ⋅ L − 1 ⋅ s − 1 , d

LuckyWell [14K]

Answer:

The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.

Explanation:

Given:

The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1

The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1

Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1

3 0

3 years ago

How long does it take Johnny to reach a speed of 80 m/s in his car while accelerating at 10m/s per second?

Rainbow [258]

Answer:

8 senkintss

Explanation:

4 0

3 years ago

What happens to the ratio when an excess quantity of a reactant exists?

In-s [12.5K]

The ratio will remain the same and the and the extra reactant will just be left.

Hope I helped !

7 0

4 years ago

A 0.145 kg baseball is thrown with a velocity of 25.0 m/s. How much work was done on the baseball to bring it from rest to 25.0

sergey [27]

Answer:

45.31 J

Explanation:

We are given that

Mass of baseball , m=0.145 kg

Initial velocity, u=0

Final velocity, v=25 m/s

We have to find the work done on the baseball to bring it from rest to 25 m/s

We know that

Work done = Change in kinetic energy

Work done, W= $\frac{1}{2}m(v^2-u^2)$

Using the formula

Work done, W $=\frac{1}{2}(0.145)((25)^2-0)$

Work done= $\frac{1}{2}(0.145)(625)$

Work done, W=45.31 J

Hence, the work done on the baseball to bring it from rest to 25 m/s

=45.31 J

3 0

3 years ago

A 100-kg astronaut floating in space holds a 10-kg object. Then the object is tossed at 50 m/s. How fast does the astronaut go a

zlopas [31]

Answer:

velocity is 5 m/s

Explanation:

mass of astronaut (M)=100kg

mass of object (m)=10kg

Initially they are rest , so initial velocity (u)= 0 m/s

Final velocity of object ; $v_1=10m/s$

let final velocity of astronaut = $v_2$ (Unknown)

Now , By conservation of linear momentum

Initial momentum = Final momentum

$(m + M)\times u = (m\times v_1) + (M\times v_2)$

Now insert values

$0 = (10 kg \times 50 m/s ) + (100 kg \: v_2)$

Further simplifying

$v_2=-\frac{500 kg-m/s}{100kg}$

So,

$v_2= -5\:m/s$

So, velocity is 5 m/s (the negative sign is due to direction)

4 0

3 years ago