Let the mass of 2500 kg car be 
and it's velocity be 
and the mass of 1500 kg car be 
and it's velocity be 
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have
2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s
Answer:
Average speed = distance/time
From 1 to 9 seconds:
Distance covered = 1 - 0.2 = 0.8 km
Time = 9 - 1 = 8 sec
Average speed = 0.8 km / 8 sec
Average speed = 0.1 km/s .
The average speed for the whole test is 1.6 km / 20 sec = 0.08 km/sec. A graph of speed vs time would average out as a horizontal line at 0.08 km/sec from 1 sec to 21 sec. The area under it would be (0.08 km/s) x (20 sec) = 1.6 km.
Surprise surprise ! The area under a speed/time graph is the distance covered during that time !
In closing, I want to express my gratitude for the gracious bounty of 3 points with which I have been showered. Moreover, the green breadcrust and tepid cloudy water have also been refreshing.
Explanation:
Answer:
Explanation:
If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .
So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .
v = u + a t
v₀ = 0 + a tmax
tmax = v₀ / a
The value of tmax is v₀ / a .
In Electrostatics the electrical force between Two charged objects is inversely Related to the distance of separation between the two objects .
Answer:
La expansión no es más que el incremento con el tiempo de la distancia entre cualquier par de galaxias lejanas. Se suele utilizar para representar este hecho la analogía de un globo donde hemos pintado una serie de puntos a modo de galaxias.
Explanation:
