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
Density I'm not sure
volume unchanged
mass unchanged
shape- water
All living organisms need energy grow and reproduce maintain their structures, and respond to their enviorments: metabolism is the set of the processes that make energy available for cellular processes.
The acceleration of gravity on or near the surface of the Earth is 9.8 m/s².
Anything acted on only by gravity loses 9.8 m/s of upward speed, or gains
9.8 m/s of downward speed, every second.
Leaping straight upward at 1.8 m/s, Tina keeps rising until she runs out of
upward speed. That happens in (1.8/9.8) = 0.1837 second after the leap.
After that, Finkel's First Law of Motion takes over:
"What goes up must come down."
The dropping part of the leap is symmetrical with the first. Please don't
make me go through proving it. Tina hits the floor at the same speed of
1.8 m/s with which she left it, and it takes the same amount of time to drop
from the peak to the floor as it took to rise from the floor to the peak.
So her total time out of contact with the floor is
2 x (0.1837 sec) = 0.367 second (rounded)
Explanation:
The given data is as follows.
F = 3.2 N, m = 18.2 kg,
t = 0.82 sec
(a) Formula for impulse is as follows.
I = Ft =
Ft =
or,
Putting the given values into the above formula as follows.
=
= 0.144 m/s
Therefore, final velocity of the mass if it is initially at rest is 0.144 m/s.
(b) When velocity is 1.85 m/s to the left then, final velocity of the mass will be calculated as follows.
Ft =
or,
=
= -1.705 m/s
Hence, we can conclude that the final velocity of the mass if it is initially moving along the x-axis with a velocity of 1.85 m/s to the left is 1.705 m/s towards the left.