The centripetal force is:
F = mv² / R
Where:
m: mass of the object
v: object speed
R: radius of the curve.
We have to:
m = 2000kg
v = 25 m / s
R = 80 meters.
Then the centripetal force acting on the vehicle is:
F = (2000kg * (25m / s) ²) / 80m
F = 15625 N
Answer:
47 m
Explanation:
Data obtained from the question include the following:
Length of dry leg 1 (L1) = 40 m
Length of dry leg 2 (L2) = 25 m
Length of swimming course (L) =..?
The length of the swimming course can be obtained by using pythagoras theory as shown below:
L² = L1² + L2²
L² = 40² + 25²
L² = 1600 + 625
L² = 2225
Take the square root of both side.
L = √2225
L = 47.1 ≈ 47 m
Therefore, the length of the swimming course is approximately 47 m.
Answer:
1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo
Explanation:
The given parameters are;
The mass of Christian and his bicycle = 54 kg
The mass of the cargo = 12 kg
1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity
∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo.
Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:
We must consider that it's launched from the ground (
) and from rest (
), with an upwards acceleration 
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:
And the velocity achieved in part 1:
We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (
) and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration 
. For the data we have it's faster to use the formula 
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:
Then, to get 
, we do:
And we substitute the values:
