Answer:
230.26 N
Explanation:
Since the speed is constant, acceleration is zero hence the net force will be given by the product of mass, coefficient of friction and acceleration due to gravity
F=0.72*32.6*9.81=230.26 N
Answer 1) : 62.5 km/hour is the average velocity of the train.
2) The final velocity of the car at the end of 75 m is 14.69 m/s
Explanation:
1) Displacement of the train = 100 km + 150 km = 250 km
Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours
Average velocity=
62.5 km/hour is the average velocity of the train.
2) The acceleration of the car, a= 1.2 
Distance covered by the car,s = 75 m
Initial velocity of the car ,
= 6 m/s
Final velocity of thre car ,
=?
Using third equation of motion:
The final velocity of the car at the end of 75 m is 14.69 m/s
Answer:
A type of telescope that does not require darkness in order to be able to use it is the refracting telescope
Explanation:
A refracting telescope consists of a lens and an eyepiece collects light which is then focused to present a magnified, bright and clear image.
The incident light on a refracting telescope is bent by refraction such that the light is focused to the focal point.
In refracting telescopes, the image is formed by bending light, that is by refraction.
The refracting telescope technology has been applied to binoculars and camera zoom lenses.
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
