Answer:
The centripetal acceleration at highway speed is greater.
Explanation:
We assume the motion of the car is uniformly accelerated. Let the highway speed be v.
By the equation of motion,
u is the initial velocity, a is acceleration and t is time
Because the car starts from rest, u = 0.
This is the tangential acceleration of the thread of the tire.
The centripetal acceleration is given by
r is the radius of the tire.
Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about
The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.
Answer:
Joe & Bo's car: 200.0 kg + 100.0 kg = 300.0 kg; Melinda's car: 25.0 kg + 100.0 kg = 125.0 kg
so the answer is 125.0kg
The solution for this problem is:
Remember that this doesn’t depend on the mass of the child.
E = T + U = constant
E (maximum height) = T + U =U = mgh = mg[r - r· cos (Θ)]
E (bottom height) = T + U = T = ½mv² = mg[r - r · cos (Θ)]
v² = 2g[r – r · cos (Θ)]
v = √ (2g[r-r·cos(Θ)])
= √(2(9.8)[3 – 3 · cos (45°)])
= 4.15 m/s or 15 kph
Answer:
1.26 secs.
Explanation:
The following data were obtained from the question:
Force (F) = 20 N
Extention (e) = 0.2 m
Mass (m) = 4 Kg
Period (T) =.?
Next, we shall determine the spring constant, K for spring.
The spring constant, K can be obtained as follow:
Force (F) = 20 N
Extention (e) = 0.2 m
Spring constant (K) =..?
F = Ke
20 = K x 0.2
Divide both side by 0.2
K = 20/0.2
K = 100 N/m
Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:
Mass (m) = 4 Kg
Spring constant (K) = 100 N/m
Period (T) =..?
T = 2π√(m/K)
T = 2π√(4/100)
T = 2π x √(0.04)
T = 2π x 0.2
T = 1.26 secs.
Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.
