The trains take <u>57.4 s</u> to pass each other.
Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.
therefore,
The relative velocity of the train A with respect to B is given by,
If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.
The trains are a distance d = 2.71 km apart.
Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,
Substitute 2.71 km for d and 170 km/h for 
Express the time in seconds.
Thus, the trains cross each other in <u>57.4 s</u>.
Alpha particles, because they are the heaviest ones (helium nuclei) and will travel around the body.
Answer:
Explanation:
The work done on a particle by external forces is defined as:
According to Newton's second law 
. Thus:
Acceleration is defined as the derivative of the speed with respect to time:
Speed is defined as the derivative of the position with respect to time:
Kinetic energy is defined as 
:
Answer:
4.0 m/s
Explanation:
The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.
Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is
where here we have
d = 3.0 m is the horizontal distance covered
vx is the horizontal velocity
t = 1.3 s is the duration of the fall
Solving for vx,
Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by
where
h = 4.0 m is the initial height
vy is the initial vertical velocity
We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy
So now we can find the magnitude of the initial velocity:
Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-
)
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
