10km/10min is a legitimate speed. So is meters/sec, km/hour (kph), etc.
Kph is very common for vehicles:
10 km/10 min (60 min/hr) = 60 kph
Since everything in the circuit is in series .. .
-- The total resistance is (3 + 2) = 5 ohms.
-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.
-- The voltage across the 2-ohm resistor is 2/5 of the total voltage.
(2/5) of (9 volts) = 18/5 = 3.6 volts .
Answer:
Examples of man-made objects that spread an impulse over a large amount of time are "airbags" in vehicles and "arrestor beds" (for trucks).
Explanation:
The question above is highly related to the topic about "Impulse" in Physics.
"Impulse"<em> refers to an object's change in momentum (the amount of motion in an object) when a force acts upon it for an interval time.</em> When it comes to providing safety to people when it comes to vehicular crashes, impulse plays a vital role.
Let's take the example of airbags in vehicles. Once a vehicle collides with another object, the driver is carried by a forward motion. Without airbags, the time is normally shorter for the driver to be stopped by the windshield. This results to a greater force. However, with the presence of air-bags, the driver will hit the airbag, instead of the windshield. <u>This will lengthen the time of the impact, thus reducing the force.</u>
Another example are the arrestor beds for trucks. Arrestor beds have been designed in order for trucks to stop, since it's hard to maneuver them. <u>With the help of arrestor beds, trucks are able to come to a stop with a longer time interval, but decreased force.</u>
work is distance * force so 15*100=1500
and to find time you know power = diastance * force / time
so 25=15*100/t
25=1500/t
25/1500=t
.016=time
Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:
lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:
To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):
=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
