Answer:
D(11) = 37.660 m
dD/dt = 2.7260 m/s
Explanation:
given data
two path apart = 17 m
walks east one path = 4 km/h = 1.111 m/s
walks west other path = 7 km/h = 1.944 m/s
pass each other time t = 0
solution
we consider here east is the positive direction and west is the negative direction
so that
the east - west distance between them is = 1.111 + 1.944 = 3.055 m/s
and
the actual distance between them time t is
D(t) =
at time 11 s
D(11) =
D(11) = 37.660 m
and
increase rate is dD/dt
dD/dt =
so for 11 sec
dD/dt =
dD/dt = 2.7260 m/s
Answer:
For each 1.5 cm the input piston moves, then output piston will move 0.5 cm
Explanation:
Let cross-sectional area of the input piston = A
Let cross-sectional area of the output piston = 3A
pressure (P) = Force (F) / Area (A)
F = PA
For a constant gravitational force on the inlet and outlet piston;
P₁A₁ = P₂A₂
But pressure = ρgh
where;
ρ is density of water
g is acceleration due to gravity
h is the distance or height moved by the piston
(ρg)h₁A₁ = (ρg)h₂A₂
h₁A₁ = h₂A₂
Area of output piston = 3 times area of input piston
h₁A₁ = h₂(3A₁)
For each 1.5 cm the input piston moves, then output piston will move;
1.5A₁ = h₂(3A₁)
1.5 = 3h₂
h₂ = 1.5 / 3
h₂ = 0.5 cm
Thus, for each 1.5 cm the input piston moves, then output piston will move 0.5 cm
The answer is True. Acceleration is a vector quantity.
Use the kinematic equation: Vf=Vi+at
Then plug;
Vi=14 m/s
a=5 m/s²
t=20 s. Therefore;
Vf=14+(5*20)
Vf=114 m/s.