"Fig is attacted with answer"
Answer:
a) d = 33.72 m
b)
= 26 m/s
c) β = 71.08°
Explanation:
a)
When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.
Given data:
time = t = 4.00 s
Height = h = 20 m
Angle = θ = 60°
Horizontal distance = d = ?
Using 2nd equation of motion

-20 =
(4) + 0.5(-9.8)(4)²
(4) = 58.4
= 14.6 m/s
This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use
=
/ sinθ
= 14.6 / sin 60
= 16.86 m/s
=
cosθ
= 16.86 cos 60
= 8.43 m/s
To calculate the horizontal distance
d =
t
d = (8.43)(4)
d = 33.72 m
b)
We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.
So,
= 
but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.
So,
g = (
-
) / t
9.8 = 14.6 -
) / 4
= 24.6 m/s
= 
= 
= 26 m/s
c)
cos β =
/ 
β = cos⁻¹ (8.43 / 26)
β = 71.08°
A substance changes from liquid to gas
Answer:

Explanation:
The acceleration of a circular motion is given by

where
is the angular velocity and
is the radius.
Angular velocity is related to the period, T, by

Substitute into the previous formula.


This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.
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