Answer:
A) 3.13 m/s
B) 5.34 N
C) W = 26.9 J
Explanation:
We are told that the position as a function of time is given by;
x(t) = αt² + βt³
Where;
α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³
Thus;
x(t) = 0.21t² + 0.0204t³
A) Velocity is gotten from the derivative of the displacement.
Thus;
v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)
v(t) = 0.42t + 0.0612t²
v(4.5) = 0.42(4.5) + 0.0612(4.5)²
v(4.5) = 3.1293 m/s ≈ 3.13 m/s
B) acceleration is gotten from the derivative of the velocity
a(t) = v'(t) = 0.42 + 2(0.0612t)
a(4.5) = 0.42 + 2(0.0612 × 4.5)
a(4.5) = 0.9708 m/s²
Force = ma = 5.5 × 0.9708
F = 5.3394 N ≈ 5.34 N
C) Since no friction, work done is kinetic energy.
Thus;
W = ½mv²
W = ½ × 5.5 × 3.1293²
W = 26.9 J
Answer:
The value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8 
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion
=
- 2 a s
Put all the values in above formula we get,
⇒ 0 =
- 2 × a ×0.2667
⇒ a = 1.2 
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.
Answer:
Beta 17,000K, bc warmer is more blue
Answer:
d = 11.1 m
Explanation:
Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

Plugging this into the energy conservation equation and cancelling m, we get

Solving for d,
