1st derivative gives velocity;
d r(t)/ dt = 2t i + 6 j + 4/t k
2nd derivative gives acceleration;
d^2 r(t)/ dt^2 = 2 i - 4/ t^2
Speed ;
Square root of (4 t^2 + 36 + 16/ t^2)
For a given time, like 2 seconds, t will be 2. And answer of speed will be scalar.
Answer:
DU = 120 Joules
Explanation:
Given the following data;
Quantity of energy = 200 J
Work = 80 J
To find the change in internal energy;
Mathematically, the change in internal energy of a system is given by the formula;
DU = Q - W
Where;
DU is the change in internal energy.
Q is the quantity of energy.
W is the work done.
Substituting into the formula, we have;
DU = 200 - 80
DU = 120 Joules
52800000000000000000000000000000000000000000
Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions
It takes twice the force to produce the same acceleration in the 2kg rock.
