Answer:
The velocity of the train is 82.8 km/h
Explanation:
The equation for the position of the train and the car is as follows:
x = x0 + v · t
Where:
x = position at time "t".
x0 = initial position.
v = velocity.
t = time.
First, let´s calculate the distance traveled by the car in 60 s (1/60 h). Let´s place the origin of the frame of reference at the front of the train when it starts to pass the car so that the initial position of the car is 0 (x0 = 0 m):
x = 0 m + 72 km/h · (1/60) h
x = 1.2 km.
Then, if the whole train passes the car at that time, the position of the front of the train at that time will be 1.2 km + 0.18 km = 1.38 km.
Then using the equation of position we can obtain the velocity:
x = x0 + v · t
1.38 km = 0 m + v · (1/60) h
1.38 km / (1/60) h = v
v = 82.8 km/h
The velocity of the train is 82,8 km/h
The same result could be obtained using the rear of the train. You only have to identify where the rear is at t = 0 and where it is at t = 60 s.
Try it!
Answer:
The Moon revolves around the Earth and rotates on its axis with the same period. • The combined effect of these two motions means that one side of the Moon always faces the Earth.
Answer:
light waves can travel in vacuum but sound waves require a medium to travel
also light waves are electromagnetic waves while sound waves are mechanical, light waves are transverse waves while sound waves are longitudinal waves
<span>Definition: The action of dividing or splitting something into two or more parts.
Reaction Example: </span><span>Autotomy, sometimes termed transverse </span>fission<span>, is the name given to a process of unequal </span>fission<span> in which a portion of the body separates off with subsequent regeneration.</span><span>
</span>
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
