Answer:
<h3>
2.3125m/s²</h3>
Explanation:
Using the equation of motion v² = u²+2aS
v is the final velocity = 120km/hr
120km/hr = 120 * 1000/1 * 3600 = 33.3m/s
u is the initial velocity = 0m/s
a is the acceleration
S is the distance covered = 240m
On substituting the given parameters
33.3² = 0²+2a(240)
33.3² = 480a
1110 = 480a
a = 1110/480
a = 2.3125m/s²
Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²
Answer:
The total resistance is 106 Ω and the current in the circuit is 0.11 A.
Explanation:
Given that,
Voltage of the battery, V = 12 V
Resistors 34 Ω, 42Ω, and 30Ω are connected in series.
The total resistance is given by :}
R = 34 + 42 +30
= 106Ω
Let I is the total current in the circuit. Using ohm's law to find it such that,
Hence, the total resistance is 106 Ω and the current in the circuit is 0.11 A.
Answer:
<h2>0.9 Joules</h2>
Explanation:
Elastic potential energy of a spring= 1/2 × Spring constant × displacement²
following calculations you will get ur answer!!
Answer:
v1 = a t + v0 = 2.5 * 5 + 5 = 17.5 m/s after 5 sec
v2 = o * 6 + 17.5 = 17.5 m/s after 11 sec
S1 = V0 t + 1/2 a t^2 = 5 * 5 + 1.25 * 5^2 = 56.25 m
S2 = 17.5 * 6 = 105 m
S = 56.3 + 105 = 161 m
Vav = S / t = 161 m / 11 s = 14.6 m/s
