An initially stationary 2.7 kg object accelerates horizontally and uniformly to a speed of 13 m/s in 4.0 s. (a) In that 4.0 s in
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Answer:
(a). The initial rate is 2.60 A/s.
(b). The rate of current increases is 0.8658 A/s.
(c). The current is 0.435 A.
(d). The final steady-state current is 0.75 A.
Explanation:
Given that,
Inductance = 2.30 H
Resistance = 8.00 Ω
Voltage = 6.00 V
(a). We need to calculate the initial rate of increase of current in the circuit.
Using formula of initial rate
Put the value into the formula
The initial rate is 2.60 A/s.
(b). We need to calculate the rate of increase of current at the instant when the current is 0.500
Using formula of rate of increase of current
....(I)
Where,
Using formula of current
The rate of current increases is
Put the value in the equation (I)
The rate of current increases is 0.8658 A/s.
(c). We need to calculate the current
Using formula of current
Put the value into the formula
The current is 0.435 A.
(d). We need to calculate the final steady-state current
Using formula of steady state
The final steady-state current is 0.75 A.
Hence, This is the required solution.
Answer:
a) 112.5 m
b) 15.81s
Explanation:
a)We can use the following equation of motion to calculate the velocity v of the rocket at s = 500 m at a constant acceleration of a = 2.25 m/s2
After the engine failure, the rocket is subjected to a constant deceleration of g = -10 m/s2 until it reaches its maximum height where speed is 0. Again if we use the same equation of motion we can calculate the vertical distance h traveled by the rocket after engine failure
So the maximum height that the rocket could reach is 112.5 + 500 = 612.5 m
b) Using ground as base 0 reference, we have the following equation of motion in term of time when the rocket loses its engine:
t = 15.81 or t = -6.33
Since t can only be positive we will pick t = 15.81s