a = ( V2 - V1)/( t2 - t1)
3.2 = ( 23.5m/s - 15.2m/s)/(t - 0)
3.2m/s = 8.3/t
t(3.2) = 8.3
t = 8.3/3.2
t = 2.59 seconds
In science, a broad idea that has been repeatedly verified so as to give scientists great confidence that it represents reality is called "a theory".
<u>Explanation:</u>
In science, the interpretation of a feature of the organic world that can be tested in repeat manner and analysed by applying agreed tests validation methods, calculation and observation in according to the scientific method, such process is called as a theory in science.
The difference lie between a theory and a hypothesis. Because hypothesis is an "educated guess". Overall it is either a proposed interpretation of an observed phenomenon, or a logical inference of a possible causal association between several phenomena.
<span>1/3
The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r"
The equation for kinetic energy is
E = 1/2MV^2.
So the energy for the system prior to collision is
0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5
The energy after the collision is
0.5rv^2
Setting the two equations equal to each other
0.5r + 0.5 = 0.5rv^2
r + 1 = rv^2
(r + 1)/r = v^2
sqrt((r + 1)/r) = v
The momentum prior to collision is
-1r + 1
Momentum after collision is
rv
Setting the equations equal to each other
rv = -1r + 1
rv +1r = 1
r(v+1) = 1
Now we have 2 equations with 2 unknowns.
sqrt((r + 1)/r) = v
r(v+1) = 1
Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r.
r(sqrt((r + 1)/r)+1) = 1
r*sqrt((r + 1)/r) + r = 1
r*sqrt(1+1/r) + r = 1
r*sqrt(1+1/r) = 1 - r
r^2*(1+1/r) = 1 - 2r + r^2
r^2 + r = 1 - 2r + r^2
r = 1 - 2r
3r = 1
r = 1/3
So the less massive particle is 1/3 the mass of the more massive particle.</span>
Answer:
53.64 m/s
Explanation:
Applying,
a = (v-u)/t............. Equation 1
Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
make u the subject of the equation
u = v-at............. Equation 2
From the question,
Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)
Substitute these values into equation 2
u = 0-(-5.364×10)
u = 0+53.64
u = 53.64 m/s
