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hichkok12 [17]
3 years ago
7

Derive the kinetic equations for Vmax and KM using the transit time and net rate constant method as discussed in class and follo

wing the attached handout. (Hint: solve for ET/v by summing the inverse net rate constants, then invert to obtain v/ET and put it in the form of kcatS/[KM S]). Remember that if a step is irreversible, the net rate constant for that step is just the rate constant for that step.

Physics
1 answer:
Pepsi [2]3 years ago
6 0

Answer:

Rate = vmax k3/k2+k3

Explanation:

The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, is referred to as Vmax.

This is usually expressed as the Km ie. Michaelis constant of the enzyme, an inverse measure of affinity. For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax.

Please kindly check attachment for the step by step solution of the given problem.

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Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
3 years ago
Hooke’s law describes the linear relationship between stress and strain through Young’s modulus. Given two materials under the s
stiks02 [169]

Answer:

The material with higher modulus will stretch less than

The material with lower modulus

Explanation:

A material with a higher modulus is stiffer and has better resistance to deformation. The modulus is defined as the force per unit area required to produce a deformation or in other words the ratio of stress to strain.

E= stress/stain

Hooks law states that provided the elastic limit is not exceeded the extension e of a spring is directly proportional to the load or force attached

F=ke

Where k is the constant which gives the measure of the spring under tension

3 0
3 years ago
Read 2 more answers
A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
tekilochka [14]
Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
A 65.8-kg person throws a 0.0413 kg snowball forward with a ground speed of 32.5 m/s. A second person, with a mass of 58.7 kg, c
guapka [62]

Answer:

v_{1} = 2.490\,\frac{m}{s}

v_{2} = 0.023\,\frac{m}{s}

Explanation:

The statement is described physically by means of the Principle of Momentum Conservation. Let assume that first person moves in the positive direction:

First Person

(65.8\,kg)\cdot (2.51\,\frac{m}{s}) = (65.8\,kg)\cdot v_{1} + (0.0413\,kg)\cdot (32.5\,\frac{m}{s} )

Second Person

(0.0413\,kg)\cdot (32.5\,\frac{m}{s})+(58.7\,kg)\cdot (0\,\frac{m}{s})=(0.0413\,kg+58.7\,kg)\cdot v_{2}

The final velocities of the two people after the snowball is exchanged is:

v_{1} = 2.490\,\frac{m}{s}

v_{2} = 0.023\,\frac{m}{s}

6 0
3 years ago
A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months
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Mem me e m even have. Jags. Shah. Shiv side esicjm is n meh dish so do indbbd
4 0
2 years ago
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