Answer:
Explanation:
height of the building h = 15 m
initial vertical velocity u = 0
v² = u² + 2 g h
u = 0 , final velocity after falling height h
h = 15
v² = 2 x 9.8 x 15
v = 17.14 m /s
. Time of fall be t
h = 1/2 g t²
15 = 1/2 x 9.8 x t²
t = 1.749 s
This time will be required for truck to come under the building .
speed = distance / time
distance = speed x time
= 22 x 1.749
38.5 m
Truck should be at a distance of 38.5 m . when the stuntman starts falling .
Answer:
as you increase in altitude, air pressure decreases
a) Hooke's law:
F = kΔx
F = spring force, k = spring constant, Δx = change of spring length
Given values:
F = 30N, Δx = 0.1m
Plug in and solve for k:
30 = k(0.1)
k = 300N/m
b) Apply the work-energy theorem; whatever work you put into deforming the spring becomes stored as spring potential energy:
W = 0.5kΔx²
W = work, k = spring constant, Δx = change of spring length
Given values:
k = 300N/m (from part a), Δx = 0.5m
Plug in and solve for W:
W = 0.5(300)(0.5)²
W = 37.5J
c) Apply the work-energy theorem here:
W = 0.5kΔx²
Given values:
k = 300N/m (from part a), Δx = 0.4m
Plug in and solve for W:
W = 0.5(300)(0.4)²
W = 24J
d) To find how much additional work needs to be done to stretch the spring an additional 0.1m if it's already been stretched 0.1m, find the potential energies for when the spring is stretched 0.2m and 0.1m and subtract them:
W = 0.5kΔx₂² - 0.5kΔx₁² = 0.5k(Δx₂² - Δx₁²)
Given values:
k = 300N/m, Δx₂ = 0.2m, Δx₁ = 0.1m
Plug in and solve for W:
W = 0.5(300)(0.2² - 0.1²)
W = 4.5J
Answer:
What is the elapsed time for Trial B? - 3.0
What is the average speed for Trial B? - 1.3
Explanation:
Answer:
un electron y un neutron se atraen o repelen: repelen
un proton y un neutron se atraen o repelen: atraen
Explanation:
