Question 26 suppose that a constant force is applied to an object. newton's second law of motion states that the acceleration of
1 answer:
You might be interested in
Answer:
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Explanation:given values
Half life of lipase t_1/2 = 8 min x 60s/min = 480 s
Rate constant for first order reaction
k_d = 0.6932/480 = 1.44 x 10^-3 s-1
Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L
rate of hydrolysis Vm0 = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3
K_m = 5mmol/L
time take is given by
all values are given and putting these value we get
t=1642.83 secs
which is equal to
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Hi there!
A.
Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.
Thus, the time to its highest point:
Now, we can determine the velocity at which the can was launched at using the following equation:
In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.
Therefore:
***vsinθ is the vertical component of the velocity.
Solve for 'v':
Now, recall that:
Plug in the expression for velocity:
B.
We can use the same process as above, where T' = 2T and Th = T.
C.
The work done in part B is 4 times greater than the work done in part A.