Answer:
9.81 × 10 = 98.1 meters
vertical displacement is s=1/2 at^2 + vt
initial vertical velocity is 0 so s=1/2 at^2
a in this instance is gravitational acceleration so 60m= 1/2 (9.81)t^2
solve for t, t = 3.497s. //I corrected this answer as just now I misread horizontal as vertical.
I think the answer is A) Cindy biked westward at 30km/h. I'm sorry if I'm wrong. If that's not correct then it's definitely C) A dog runs an average of 8mp/h. I hope this helps. :)
Answer:
Average :
UCL = 4.15
LCL = 2.65
Range :
UCL = 2.75
LCL = 0
Explanation:
Given :
Sample size, n = 5
Average, X = 3.4
Range, R = 1.3
A2 for n = 5 ; equals 0.577 ( X chart table)
For the average :
Upper Control Limit (UCL) :
X + A2*R
3.4 + 0.577(1.3) = 4.1501
Lower Control Limit (LCL) :
X - A2*R
3.4 - 0.577(1.3) = 2.6499
FOR the range :
Upper Control Limit (UCL) :
UCL = D4*R
D4 for n = 5 ; equals = 2.114
UCL = 2.114*1.3 = 2.7482
Lower Control Limit (LCL) :
LCL = D3*R
D3 for n = 5 ; equals = 0
LCL = 0 * 1.3 = 0
The distance covered by car is equal to (assuming it is moving by uniform motion) the product between the car's speed and the time of the car ride, 4 h:

where

is the car's speed

is the duration of the car ride
Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:

The total distance covered is S=255 km, which is the sum of the distances covered by car and train:

which becomes

(1)
we also know that the train speed is 5 km/h greater than the car's speed:

(2)
If we put (2) into (1), we find

and if we solve it, we find


So, the car speed is 20 km/h and the train speed is 25 km/h.
Answer:
Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.