Answer:
<h2>The answer is 4 s</h2>Explanation:
The time taken can be found by using the formula
where
d is the distance
v is the velocity
From the question we have
We have the final answer as
<h3>4 s</h3>Hope this helps you
With the switch closed, how does the voltage across the 20-ω resistor compare to the voltage across the 10-ω resistor?
1 answer:
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Answer:
T = 295.57 s
Explanation:
given,
mass of the rocket = 200 Kg
mass of the fuel = 100 Kg
acceleration = 35 m/s²
time, t = 35 s
time taken by the rocket to hit the ground, = ?
Final speed of the rocket when fuel is empty
using equation of motion
v = u + a t
v = 0 + 35 x 35
v = 1225 m/s
height of the rocket where fuel is empty
v² = u² + 2 a s
1225² = 0 + 2 x 35 x h₁
h₁ = 21437.5 m
After 35 s the rocket will be moving upward till the final velocity becomes zero.
Now, using equation of motion to find the height after 35 s
v² = u² + 2 g h₂
0² = 1225² + 2 x (-9.8) h₂
h₂ = 76562.5 m
total height = h₁ + h₂
= 76562.5 m + 21437.5 m = 98000 m
now, time taken by before the rocket hit the ground
using equation of motion
negative sign is used because the distance travel by the rocket is downward.
4.9 t² - 1225 t - 13500 = 0
t = 260.57 s
neglecting the negative sign
total time the rocket was in air
T = t₁ + t₂
T = 35 + 260.57
T = 295.57 s
Time for which rocket was in air is equal to 295.57 s.
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
