What is the mass of 1.0 l of water in grams?

The location where an earthquake begins

Ivahew [28]

Answer:

b

Explanation:

the location where an earthquake begin is called

3 0

3 years ago

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

Pepsi [2]

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.

6 0

3 years ago

Heat is applied to an ice cube in a closed container until only steam is present. draw a representation of this process, assumin

olga55 [171]

The solid, liquid and gas phases of water would have the same structure of the molecules since they are same substance. The only difference would be the distances of the molecules in the container. For a ice, the molecules are close to each other where the molecules vibrate only in place. For liquid, the molecules are freely moving and are at some distance with each other but not that far away with each other. Steam, on the other hand, would have molecules that are very far from each other and are freely moving in the whole container. As the container is heated, the size of the molecules would not change. It is only the volume that has changed. Also, the mass is the same since there is no outflow of the substances.

6 0

4 years ago

A uniform 2.50m ladder of mass 7.30kg is leaning against a vertical wall while making an angle of 51.0degree with the floor. A w

Zigmanuir [339]

Answer:

19.95 J

Explanation:

The center of mass of the ladder is initially at a height of:

$h_1=\frac{L}{2}sin\theta$

The center of mass of the ladder ends at a height of:

$h_2= \frac{L}{2}sin90$ =L/2

So, the work done is equal to the change in potential energy which is:

W = PE = $mg(h_2-h_1)$

now $h_2-h_1= 1-sin\theta$

therefore

W = [mgL/2]×[1 - sin(theta)]

W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]

solving this we get

W = 19.95 J

8 0

3 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$