Answer:
0.9432 m/s
Explanation:
We are given;
Mass of swimmer;m_s = 64.38 kg
Mass of log; m_l = 237 kg
Velocity of swimmer; v_s = 3.472 m/s
Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.
So;
Initial momentum = final momentum
m_l × v_l = m_s × v_s
Where v_l is speed of the log relative to water
Making v_l the subject, we have;
v_l = (m_s × v_s)/m_l
Plugging in the relevant values, we have;
v_l = (64.38 × 3.472)/237
v_l = 0.9432 m/s
Also 4 sec. Without air resistance, all objects fall with the same acceleration, and reach bottom in the same amount of time.
Answer:
1.15*10^-5 kg/m^3
Explanation:
Given data
mass= 1kg
hieght= 40 mm
diameter= 52.5mm
radius= 53.5/2= 26.25mm
The volume of the cylinder
V=πr^2h
V=3.142*26.25^2*40
V=3.142*689.0625*40
V=86601.375 mm^3
Density= mass/volume
Density= 1/86601.375
Density=0.00001154716
Density= 1.15*10^-5 kg/m^3
Hence the density is 1.15*10^-5 kg/m^3
Answer:
(a) the tangential speed of a point at the edge is 3.14 m/s
(b) At a point halfway to the center of the disc, tangential speed is 1.571 m/s
Explanation:
Given;
angular speed of the disc, ω = 500 rev/min
diameter of the disc, 120 mm
radius of the disc, r = 60 mm = 0.06 m
(a) the tangential speed of a point at the edge is calculated as follows;
Tangential speed, v = ωr
v = 52.37 rad/s x 0.06 m
v = 3.14 m/s
(b) at the edge of the disc, the distance of the point = radius of the disc
at half-way to the center, the distance of the point = half the radius.
r₁ = ¹/₂r = 0.5 x 0.06 m = 0.03 m
The tangential velocity, v = ωr₁
v = 52.37 rad/s x 0.03 m
v = 1.571 m/s
