Let's consider a few possibilities.
1. The lowest velocity of the paratrooper would be just before hitting the ground.
2. Given that the jump originated from a relatively short height, the paratrooper utilized a static line, allowing the parachute to deploy almost instantly after leaping.
Hence, we will convert 100 mi/h to ft/s:
100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec.
Based on the first assumption, the maximum distance fallen by the paratrooper would equate to 8 seconds at 146.67 ft/s, translating to
8 s * 146.67 ft/s = 1173.36 ft.
This calculated distance is nearly on par with the jump height, validating both assumptions 1 and 2. Thus, this scenario seems plausible.
Moreover, considering the terminal velocity for a parachutist in a freefall position with limbs spread out typically reaches 120 mi/h, which is slightly above the 100 mi/h mentioned in the article. This as well aligns with the notion of the parachute acting like a flag, adding some air resistance.
Answer:
Electric flux is calculated as 
Explanation:
We start with the given parameters:
The electric field impacting the circular surface is 
Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:
Applying properties of the dot product, we calculate the electric flux as:
Consequently, the electric flux for the circular area is . Thus, this represents the required answer.
(6-16)/4.0=-2.5 m/s²
The car's acceleration is -2.5 m/s²
We know that F=ma
where m represents mass and a indicates acceleration
thus, Force= ma
therefore, F=1300X1.07=1391N
I hope this helps
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b. The electric field diverges from the source charge when it is positive.
d. If the source charge is negative, the electric field converges towards it.
In more depth:
A positive source charge generates an electric field that delivers a repulsion to a positively charged test charge. Hence, the field points outward from positively charged sources. Conversely, a positively charged test charge is drawn toward a negative source charge, resulting in electric field vectors consistently pointing towards negatively charged objects.
Additionally, the intensity of the electric field solely depends on the test charge.
Thus, the valid responses are b and d, confirming that the electric field points away from a positive source charge and toward a negative source charge.
