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: The time required for the impluse passing through each other is approximately 0.18seconds
Explanation:
Given:
Length,L = 50m
M/L = 0.020kg/m
FA = 5.7×10^2N
FB = 2.5×10^2N
The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.
Ca(t) + CB(t) = 50
Where CA and CB are the velocities of the wire A and B
t = 50/ (CA + CB)
But C = Sqrt(FL/M)
Substituting gives:
t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))
t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))
t = 50 / (168.62 + 111.83)
t = 50/280.15
t = 0.18 seconds
Answer:
Center of mass lies 24 m in front of center of mass of second wagon.
Explanation:
Suppose A 1200 kg station wagon is moving along a straight highway at 12.0 m/s. Another car with mass 1800 kg and speed 20.0 m/s.
Given that,
Mass of first wagon = 1200 kg
Mass of second wagon = 180 kg
Distance = 40 m
We need to calculate the position of the center of mass of the system
Using formula of center mass
Hence, Center of mass lies 24 m in front of center of mass of second wagon.
