The average speed between 0 h and 2.340 h is 6.97 Km/h
Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.

With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:
- Total time = 2.340 – 0 = 2.340 h
- Total distance = 16.3 – 0 = 16.3 Km
- Average speed =?

Learn more about average speed: brainly.com/question/24884027
Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
Answer:

Explanation:
Given that
Constant rate of leak =R
Mass at time T ,m=RT
At any time t
The mass = Rt
So the total mass in downward direction=(M+Rt)
Now force equation
(M+Rt) a =P- (M+Rt) g

We know that





This is the velocity of bucket at the instance when it become empty.
To solve this problem we will use a free body diagram that allows us to determine the Normal Force.
In general, the normal force would be equivalent to

Since the skier is standing on two skis, his weight will be divide by two

Pressure is given as the force applied in a given area, that is

Replacing F with N'


Our values are given as,




Replacing we have that


Therefore the pressure exerted by each ski on the snow is 776.01Pa