Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
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:
Answer:
The rate of change of distance is defined as speed.
Explanation:
The speed is defined as the rate of change of distance.
Speed = distance/ time
When we know the distance and the time, we get the value of speed. So, e know that who is moving fast or slow.
hen a graph is pltted beteen the distance and time, the slope of the graph gives the value of speed. So, by checking the slopes, hoseslope ismore, the speed is more and thusit is moving faster.
So, i agree with the statement.
Answer:
32 seconds
Explanation:
m1 = 80 kg
m2 = 10 kg
v2 = 5m/s
According to the property of conservation of momentum, assuming that both you and the bag are stationary before the safety rope comes lose:
Since the space station is 20 meters away, the time taken to reach it is given by:
It takes you 32 seconds to reach the station.
