The kinetic energy of an object is directly proportional to its mass and the square of its velocity
KE = 1/2 (mv²)
KE = Kinetic Energy
m = mass in kg
v = velocity in m/s
Given:
m = .8 kg
v = 11.2 m/s
Substitute:
KE = 1/2 (.8)(11.2²)
KE = 50.18 J
Answer: The reference frame of a passenger in a seat near the center of the train
Explanation:
the speed of light is the same for the passenger and the bicyclist
then the avents are simultaneous fo the passenger not for the bicyclist
the delay between the two events for the bicyclist is
Δt=Δd/vs
where
Δd= lenght of train
vs=speed of sound
the reference frame of a passenger in a seat near the center of the train
Solution:
The space and time transformations are:
x' = γ(x - vt)
t' = γ(t - vx/c²).
In the primed frame the two events are simultaneous, so that Δt' = 0. Also here Δx' = 30. In the unprimed frame Δx' = 30 = γ(Δx - v Δt).......(*)
We also have Δt' = 0 = γ(Δt - vΔx/c²)→Δx = c²Δt/v......(**)
Substituting (**) in (*): 30 = γ(c²Δt/v - vΔt))→Δt = 30/(c²/v - v) =
30/(2c - 0.5c) = 6.7 x 10^(-8)s
If we see the forces on the astronauts then there is only one force on each astronaut while they are in air.
This force is due to gravity of earth.
Now while the astronauts are in air and doing some activities then the net force on them is counterbalanced on them by centrifugal force.
so we can say the rate of fall of astronauts due to gravity is at same rate as the orbiting rate of the space shuttle.
this is given by

so all astronauts will experience the situation of free fall.
The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.
Answer:
Here's what I get
Explanation:
A. Distance between A and B.
h = -½gt²
The stones go faster the farther they fall.
Stone A has already reached 5 m when B is released.
When B reaches 5 m, A has dropped further and is falling even faster.
The distance between the stones increases with time.
Figure 1 shows this effect in a graph of height vs. time.
B. Speed of Stone B
v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²
v = 9.9 m/s
The stone is travelling at 9.9 m/s when it reaches 5 m.
C. Velocity vs time
v = -gt
Both stones accelerate at the same rate.
When Stone B has reached 10 m at time t, Stone A is falling much faster.
Fig. 2 shows this in a graph of velocity vs time.