Answer:
The maximum speed of the jumper is 54.2m/s
The spring constant of the rope is
K=9.19KN/m
Explanation:
Step one
According to hook's law the applied force F is proportional to the extension e provided the elasticity is maintained
Step two
Given that
mass= 50kg
Height of bridge h=150m
Extention e=4m
Step three
To determine the maximum speed of the jumper
Since he is going with gravity we assume g=9.81m/s²
And we apply the equation of motion
V²=U²+2gh
where u= initial velocity of the jumper =0
h=height of bridge
Step four
Substituting we have
V²=0²+2*9.81*150
V²=2943
V=√2943
V=54.2m/s
Step five
To solve for the spring constant
We have to equate to potential energy of the jumper to the energy stored in the spring
Potential energy of the jumper =mgh
Energy stored in the spring =1/2ke²
Hence mgh=1/2ke²
Making k subject of formula we have
K=2mgh/e²
Substituting our data into the expression we have
K=2*50*9.81*150/4²
K=147150/16
K=9196.9N/m
K=9.19KN/m
Answer:
third
According to Newton's Third Law, for every action there is an equal and opposite reaction. The player kicks the soccer ball, and the ball “kicks” back, but the player doesn't feel the reaction because the player's leg has more mass and force than the soccer ball.
Answer:
a) v = 7.69 10³ m / s, b) T = 92.6 min
Explanation:
a) For this exercise we use the centripetal acceleration ratio, which in itself assumes a circular orbit, is equal to the acceleration of gravity
a = v² / r
v = 
the distance to the ISS is
r = R_earth + d
r = 6400 10³ + 400 10³
r = 6800 10³ m
we calculate
v = 
Ra (8.69 6800 103)
v = 
v = 7.687 10³ m / s
the result with the correct significant figures
v = 7.69 10³ m / s
b) The speed of the ISS is constant, so we can use the uniform motion relationships
v = d / t
if distance is the orbit distance
d = 2π r
time is called period
v = 2π r / T
T = 2π r / v
let's calculate
T = 2π 6800 10³ /7,687 10³
T = 5.558 10³ s
let's reduce the period to minutes
T = 5.558 10³ s (1 min / 60s)
T = 9.26 10¹ min
T = 92.6 min
Answer:
3.7 m
Explanation:
ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky
The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of
t = √ (2h/g) = √(2(15)/9.8) = 1.75 s
Without the tail wind, the ball travels horizontally
d = vt = 68(1.75) = 119 m
The tailwind exerts a constant acceleration on the ball of
a = F/m = 12/5.0 = 2.4 m/s²
The average horizontal velocity during the flight is
v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s
so the distance with tailwind is
d = v(avg)t = 70.1(1.75) = 122.675 m
The extra distance is 122.675 - 119 = 3.675 = 3.7 m
