Solution :
Given weight of Kathy = 82 kg
Her speed before striking the water, 
= 5.50 m/s
Her speed after entering the water, 
= 1.1 m/s
Time = 1.65 s
Using equation of impulse,
Here, F = the force ,
dT = time interval over which the force is applied for
= 1.65 s
dP = change in momentum
dP = m x dV
![$= m \times [V_f - V_o] $](https://tex.z-dn.net/?f=%24%3D%20m%20%5Ctimes%20%5BV_f%20-%20V_o%5D%20%24)
= 82 x (1.1 - 5.5)
= -360 kg
∴ the net force acting will be
= 218 N
Answer:
cohesive properties
Explanation:
The property of cohesion allows liquid water to have <u>no tension on the surface</u>.
Use a=(dv/dt) (change in velocity/ change in time)=acceleration
(1.2/5)=acceleration
F=ma (Newton's second law, Force= Mass x Acceleration
=960 x 0.24 F=230.4N If T<230.4N then the tow rope will hold
Answer:
t_total = 23.757 s
Explanation:
This is a kinematics exercise.
Let's start by calculating the distance and has to reach the limit speed of
v = 18.8 m / s
v = v₀ + a t₁
the elevator starts with zero speed
v = a t₁
t₁ = v / a
t₁ = 18.8 / 2.40
t₁ = 7.833 s
in this time he runs
y₁ = v₀ t₁ + ½ a t₁²
y₁ = ½ a t₁²
y₁ = ½ 2.40 7.833²
y₁ = 73.627 m
This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.
x_total = x₁ + x₂
x₂ = x_total - x₁
x₂ = 373 - 73,627
x₂ = 299.373 m
this distance travels at constant speed,
v = x₂ / t₂
t₂ = x₂ / v
t₂ = 299.373 / 18.8
t₂ = 15.92 s
therefore the total travel time is
t_total = t₁ + t₂
t_total = 7.833 + 15.92
t_total = 23.757 s
