Answer:
C 350W
Explanation:
Given power output to walk on a flat ground to be 300W, h = 0.05x, v = 1.4m/s
m = 70kg and g =9.8m/s².
x = horizontal distance covered
Total energy used = potential energy used in climbing and the energy used in a walking the horizontal distance.
E = mgh + 300t
Where t is the time taken to cover the distance
x = vt and h = 0.05vt
So
E = mg×0.05×vt + 300t
Substituting respective values
E = 70×9.8×0.05×1.4t +300t = 348t
P = E/t = 348W ≈ 350W.
Consider the motion of the car before brakes are applied:
v₀ = maximum initial velocity of the car before the brakes are applied
t = reaction time = 0.50 s
x₀ = distance traveled by the car before brakes are applied
since car moves at constant speed before brakes are applied
Using the equation
x₀ = v₀ t
x₀ = v₀ (0.50)
Consider the motion after brakes are applied :
v₀ = initial velocity of the car before the brakes are applied
a = acceleration = - 10 m/s²
v = final velocity of the car after it comes to stop = 0 m/s
x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)
Using the equation
v² = v²₀ + 2 a x
inserting the values
0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))
v²₀ = 20 (38 - v₀ (0.50))
v₀ = 23 m/s
Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
At constant speed, we will have;
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
Momentum is a vector quantity, and is always conserved. Whenever a collision occurs between two objects, the objects behave under the principle of conservation of momentum. Therefore, if an object moves in the direction opposite to its original direction after a collision, then this indicates that the momentum of the colliding object was greater than the object under consideration.
