A car travelling at 5.0 m/s starts to speed up. After 3.0 s its velocity has increased to 11 m/s. a) What is its acceleration? (
1 answer:
initially, the car is traveling at 5.0 m/s.
so, we know acceleration for changing velocity is :
a = (v-v)/t ..........(i)
where v is the final velocity
v is the initial velocity
t is the time taken to change velocity
Now, as per the question :
initial velocity, v=5.0 m/s
final velocity, v =11 m/s
time taken, t = 3 s
putting the values in equation (i),
a = ( 11-5 )/3
a = 2 m/s²
Therefore, a, after 3 s, is <em>2 m/s².</em>
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Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m
Answer:
A. 9.31 x10^10J
B. -8.47x10 ^ 12J
C. 8.38x 10^12J
Explanation:
See attached file pls
