Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
Risk of return on investment is higher than other forms of energy generation.
Equations of the vertical launch:
Vf = Vo - gt
y = yo + Vo*t - gt^2 / 2
Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2
=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s
The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.
Answer: 11.25 m/s
Answer:
1. Distance travelled = 12 km.
2. Displacement = 8.6 km
Explanation:
From the question given above, the following data were obtained:
Distance 1 (d₁) = 7 km
Distance 2 (d₂) = 5 km
Total distance =?
Displacement =?
1. Determination of the distance travelled.
Distance 1 (d₁) = 7 km
Distance 2 (d₂) = 5 km
Total distance (dₜ) =?
dₜ = d₁ + d₂
dₜ = 7 + 5
dₜ = 12 km
2. Determination of the displacement.
In the attached photo, R is the displacement.
We can obtain the value of R by using the pythagoras theory as illustrated below:
R² = 7² + 5²
R² = 49 + 25
R² = 74
Take the square root of both side
R = √74
R = 8.6 km
