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valentina_108 [34]
3 years ago
9

In a certain kinetics experiment, the enzymatically catalyzed hydrolysis of ATP proceeds at a constant rate of 2.0 µM•s–1. If th

e volume of solution is 1.0 mL, what is the total number of ATP molecules that hydrolyzed after 1 min?
Physics
1 answer:
DedPeter [7]3 years ago
4 0

Answer:

Explanation:

Rate of hydrolysis = 2 x 10⁻⁶ M per second = 2 x 10⁻⁶ x 60 M per minute

2 x 10⁻⁶ x 60 x 6.03 x 10²³ molecules per minute

= 723.6 x 10¹⁷ molecules/min.

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Which of the following cart configurations could produce the X versus T graph shown
Xelga [282]

Answer:

B. Cart B

Explanation:

8 0
3 years ago
If you doubled the load resistor in a Wheatstone bridge, the load current would not be half as much. Why not?
Fofino [41]
Doubling the size of a load resistor increases the load current. Increasing the load resistance, in turn the total circuit resistance is reduced. The load current would not be half as much since when you increase the size of load resistor then load current increases.
6 0
3 years ago
The law of universal gravitation states that any two objects in the universe, without exception,
STatiana [176]
A. attract each other.

The Law of Universal Gravitation discusses the phenomenon of gravity. Remember that gravity is the force that keeps us on Earth; the Earth pulls us down, and our bodies pull back.  Gravity is the force of attraction, so the correct answer is a).
3 0
3 years ago
A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0
oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box m=2\ kg

speed of box v=1.9\ m/s

distance moved by the box x=10\ cm

coefficient of kinetic friction \mu _k=0.66

Friction  force f_r=\mu_kN

f_r=0.66\times mg

f_r=0.66\times 2\times 9.8=12.936 \N

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2

\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2

3.61-1.2936=0.005\times k

k=463.28\ N/m

3 0
3 years ago
Read 2 more answers
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
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