Answer:
speed of eight ball speed after the collision is 3.27 m/s
Explanation:
given data
initially moving v1i = 3.4 m/s
final speed is v1f = 0.94 m/s
angle = θ w.r.t. original line of motion
solution
we assume elastic collision
so here using conservation of energy
initial kinetic energy = final kinetic energy .............1
before collision kinetic energy = 0.5 × m× (v1i)²
and
after collision kinetic energy = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²
put in equation 1
0.5 × m× (v1i)² = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²
(v2f)² = (v1i)² - (v1f)²
(v2f)² = 3.4² - 0.94²
(v2f)² = 10.68
taking the square root both
v2f = 3.27 m/s
speed of eight ball speed after the collision is 3.27 m/s
you tend to push off with the foot still on the boat.
Newtons 3rd law of motion hold that every action has an equal and opposite reaction.
So, the amount of force you use on your back foot to push yourself onto the dock, has an equal end opposite amount of force going the other way from your sole of your foot, which pushes the boat the other way.
The same principle applies when you fire a cannon - in the pirate films, you see the body of the cannon forced back as it is fired.
Take it one step further, Henry VIII flagship, “Mary Rose fired all its cannons together one day in a broad side, the opposite force rolled the ship over and sank it.
Answer:
k = 0.5 MN/m
Explanation:
Mass of the railcar, m = 5000 kg
Speed of the rail car, v = 1 m/s
The Kinetic energy(KE) of the railcar is given by the equation:
KE = 0.5 mv²
KE = 0.5 * 5000 * 1²
KE = 2500 J
The spring's compression, x = 0.1 m
The potential energy(PE) stored in the spring is given by the equation:
PE = 0.5kx²
PE = 0.5 * k * 0.1²
PE = 0.005k
According to the principle of energy conservation, Kinetic energy of the railcar equals the potential energy stored in the spring
KE = PE
2500 = 0.005k
k = 2500/0.005
k = 500000 N/m
k = 0.5 MN/m
The displacement of a moving object is the straight-line distance between the place it starts from and the place where it stops.
The displacement of anything moving along a circular track depends on how far around it goes before it stops. The greatest displacement it can possibly have is the diameter of the track ... 100m on this particular one ... because that's as far apart as two places on a circle can ever be.
The most interesting case is when the object goes around the circle exactly once. Then it stops at the same place it started from, the distance between the starting point and ending point is zero, and after all that motion, the displacement is zero.
Answer:
(a). The mass of the box is 47.8 kg.
(b). The coefficient of kinetic friction between the floor and the box is 0.093.
Explanation:
Given that,
Force = 71 N
Acceleration = 0.57 m/s²
Pushing force = 82 N
Change acceleration = 0.80 m/s²
(a). We need to calculate the mass of the box
Using balance equation
Put the value in the equation
....(I)
....(II)
From equation (I) and (II)
The mass of the box is 47.8 kg.
(b). We need to calculate the friction force
Now, put the value of m in equation (I)
We need to calculate the coefficient of kinetic friction between the floor and the box
Using friction force
The coefficient of kinetic friction between the floor and the box is 0.093.
Hence, (a). The mass of the box is 47.8 kg.
(b). The coefficient of kinetic friction between the floor and the box is 0.093.
