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:
its all black I cant see the image can you re upload it
Explanation:
Answer:
distance between both oasis ( 1 and 2) is 27.83 Km
Explanation:
let d is the distance between oasis1 and oasis 2
from figure
OC = 25cos 30
OE = 25sin30
OE = CD
Therefore BC = 30-25sin30
distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem
in
PUTTING ALL VALUE IN ABOVE EQUATION
d = 27.83 Km
distance between both oasis ( 1 and 2) is 27.83 Km
Answer:
a single closed path of electrical components including a voltage source
Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math
at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
ke = work (use all ke to do work)
(1/2)mv^2 = umgh x cot(a)
u = (1/2)v^2 x tan (a) / gh
