Answer:
D = 30.625 m
Explanation:
given,
Speed of the climber = 1.3 m/s
time = 2.5 s
acceleration due to gravity = 9.8 m/s²
initial speed of the kit = 1.3 m/s
velocity of the kit after 2.5 s
using equation of motion
v = u + a t
v = 1.3 + 9.8 x 2.5
v = 25.8 m/s
distance travel by the kit in 2.5 s
v² = u² + 2 g h
25.8² = 1.3² + 2 x 9.8 x h
19.6 h = 663.95
h = 33.875 m
distance travel by the rock climber in 2.5 s
distance = speed of climber x time
h' = 1.3 x 2.5
h' = 3.25 m
Distance between kit and rock climber
D = h - h'
D = 33.875 - 3.25
D = 30.625 m
The kit is 30.625 m below climber.
Answer:
1120 N
Explanation:
The velocity with which he hits the water can be found with kinematics:
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)
v = -13.3 m/s
Or it can be found with conservation of energy.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × -9.8 m/s² × -9.00 m)
v = -13.3 m/s
Sum of forces on the diver after he hits the water:
∑F = ma
F − mg = m Δv/Δt
F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)
F = 1120 N
D. F=ma
F is for force, and that equals two things, M for mass and A for acceleration. When mass is accelerated, it gives you force. Force equals multiplying mass and its acceleration.
Answer:
5 Pascals
Explanation:
Pressure is equal to force/area and the area is 60cm to m is 0.6m. Then the area is 0.6m×5m=3m^2. which gives a pressure of 15N/3m^2=5Pa