<span>Let F represent the maximum thrust produced by the car's motor. Thus, F = ma = 1300 x 3.0 = 3900 N. After adding the load, F stays the same, leading to the equation F = 1700a, which results in a = F/1700 = 3900/1700 = 2.3 m/s².</span>
The volumes are 200cm3 and 0.0002m3
Answer: 339.148N
Explanation:
Given data:
Time (t) = 47s
Initial speed (U) = 0m/s
Final speed (V) = 9.5m/s
Mass of B = 540kg
Frictional force on B = 230N
Since both boats are linked, movement of A causes B to move as well.
What is the acceleration of boat A?
Applying the motion formula:
V = u + at
9.5 = 0 + a * 47
a = 9.5 / 47
a = 0.2021 m/s²
To determine the force necessary to accelerate boat B, as both boats experience the same force:
F = Mass * acceleration
F = 540 * 0.2021 = 109.14N
Given that there is a frictional force of 230N acting on boat B, the overall force (Tension) becomes:
Tension = frictional force + applied force = (109.14 + 230)N = 339.148N
1/0.0545. The transformation ratio of primary coil turns to secondary coil turns is directly proportional to the voltage transformation occurring. With 6.0 V on the secondary side (output) and 110 V on the primary side (input), the voltage ratio is calculated as 6/110 = 0.0545. This means for each turn in the primary coil, there are 0.0545 turns in the secondary coil.
Answer:

Explanation:
Let the charge on the large droplet be denoted as Q.
When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".
Applying volume conservation gives us:


Now, the potential for the smaller droplets is given as:



