Answer:
a) t = 1.75 s
b) x = 31.5 m
Explanation:
a) The time at which Tom should drop the net can be found using the following equation:
Where:
: is the final height = 0
y₀: is the initial height = 15 m
g: is the gravity = 9.81 m/s²
: is the initial vertical velocity of the net = 0 (it is dropped from rest)
Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.
b) We can find the distance at which is Jerry when Tom drops the net as follows:
Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.
I hope it helps you!
Answer: potential energy but no kinetic energy
Explanation:
Since the rock is stationary, velocity is zero, therefore no kinetic energy,but there's potential energy because the rock is at rest,
Answer:
Increase 9 times
Explanation:
We have Newton formula for attraction force between 2 objects with mass and a distance between them:
where 
is the gravitational constant. 
is the masses of the 2 objects. and R is the distance between them.
Since the force is inversely proportional to the distance squared, if it is reduced by 3 times, the gravitational force between them would increase by 
times
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
