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>
Answer:
we can say that wind energy is due to
D) Severe thunderstorms
Explanation:
As we know that wind energy is converted into kinetic energy of wind mills
This kinetic energy of wind mill is then converted into electrical energy using turbine
now we can consider here energy conservation theory that energy is only converted from one form to other form
it neither be destroyed nor be created but it can transfer from one form to other form
So here we can say that wind energy is due to
D) Severe thunderstorms
Answer:
50.4÷1.18= 42.7m/s as speed is equal to the distance over the time taken
The answer is 3+5+4 = 10+3 so then you have to add the number to the part of the equation and you will get the answer of five
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i ( 
)
1 = i ( 
)
1 = i ( 
) = 
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ = 
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ = 
R₂ =
R₂ = 19.9 Ω
