The answer would be to research the need. This should have been done before the project began.
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:
, where the minus indicates the direction is opposite to that of the throw.
Explanation:
a)
Since MKS stands for meter-kilogram-second and we know that:
We can write that:
These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.
So we have that:
b)
Newton's 2nd Law tells us that F=ma, and the definition of acceleration is , so we have:
Taking the throw direction as the positive one, for our values we have:
If it were possible for an object to fall freely near the surface of the Earth,
-- The direction of its velocity would always be "down"; that is, toward the center of the Earth.
-- The size of its velocity would continually increase, at the rate of 9.8 meters per second for every second it falls.
The answer is parallel
If the <span>circuits in a car</span> were series, they would go out at the same time.
I hope this helps! :3