I think it’s saturated, not completely sure
The average time that it takes for the car to travel the first 0.25m is 2.23 s
The average time that it takes for the car to travel the first 0.25 m is given by:
The average time to travel just between 0.25 m and 0.50 m is 0.90 s
First of all, we need to calculate the time the car takes in each trial to travel between 0.25 m and 0.50 m:
Then, the average time can be calculated as
Given the time taken to travel the second 0.25 m section, the velocity would be 0.28 m/s
The velocity of the car while travelling the second 0.25 m section is equal to the distance covered (0.25 m) divided by the average time (0.90 s):
Answer:
Explanation:
Magnitude of force per unit length of wire on each of wires
= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .
Putting the values ,
force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )
= .67 i² x 10⁻⁴
force on 3 m length
= 3 x .67 x 10⁻⁴ i²
Given ,
8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²
i² = 3.98 x 10⁻²
i = 1.995 x 10⁻¹
= .1995
= 0.2 A approx .
2 i = .4 A Ans .
Answer:
Explanation:
30 N force is pulling total mass of 15 kg , so acceleration in the system of masses
= 30 / 15
= 2 m / s²
Let us now consider forces acting on 9 kg . 30 N is pulling it in forward direction . Tension T in the string attached to it is pulling it in reverse direction
so net force on it
30 - T
Applying Newton's law of motion on it
30 - T = mass x acceleration
30 - T = 9 x 2
30 - 18 = T
T = 12 N
