Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Answer:
Heat capacity is the ratio of the amount of heat energy transferred to an object to the resulting increase in its temperature. Molar heat capacity is a measure of the amount of heat necessary to raise the temperature of one mole of a pure substance by one degree K.
Explanation:
hope this helps <3
Answer:
a. 1715 N b. 2401 N
Explanation:
Let F = force due to calf muscle, F' = force due to tibia and N = force due to ground = weight of man = mg where m = mass of man = 70 kg and g = acceleration due to gravity = 9.8 m/s².
a. Magnitude of the forces exerted on the foot by the muscle
Since the force due to the calf muscle is 6.0 cm behind the ankle joint and the normal force due to the ground is 15.0 cm in front of the ankle joint and the force due to the tibia is at the ankle joint, taking moments about the ankle joint,
F × 6 cm + F' × 0 cm = N × 15 cm
6F = 15N = 15mg
F = 15mg/6
= 15 × 70 kg × 9.8 m/s²/6
= 1715 N
b. Magnitude of the forces exerted on the foot by the tibia
Taking moments about the calf muscle force, we have
F × 0 cm + F' × 6 cm = N × (15 cm + 6 cm)
6F' = 21N = 21mg
F' = 21mg/6
= 21 × 70 kg × 9.8 m/s²/6
= 2401 N
Answer:
OPTIMISATION
Explanation:
Optimisation is a mathematical theory for developing strategies that maximize gains and minimize losses while adhering to a given set of rules and constraints.
The theory has a target function to be maximised or minimised, dependent on its explanatory variable(s), with respect to which the function has to be maximised or minimised. It also has constraints which might be binding factors to maximisation / minimisation.
Eg : Revenue optimising output is found by maximising profit function with respect to constraint function in forms of cost etc.
Dividing x^3 - 3x^2 - 4x + 12 by x + 2 gives
x^2 -5x + 6
= (x - 3)(x - 2)
so remaining factors are x - 3 and x - 2 answer