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:
h = height of the hotel room from the ground floor = 237.4m
Explanation:
Change in Potential Energy of tourist = ΔPE = PE2 – PE1 = mgh
PE1 is the potential energy of tourist at the ground floor
PE1 is the potential energy of tourist at the top (hotel room)
Given
PE1 = − 2.01 × 10⁵ J
PE2 = 0J
PE2 – PE1 = mgh
0 – (− 2.01 × 10⁵ J) = mgh
2.01 × 10⁵ J = 86.4×9.8×h
h = 2.01 × 10⁵/(86.4×9.8) = 237.4m
Vo= 331+0.6T
360=331+0.6T
360-331=0.6T
29=0.6T
0.6T/29
T=6/290 so change it to simplest form and us formulas good luck
Explanation:
Let us first calculate long does it take to go 12m at 30m/s( assumed speed)
12/30 = 0.4 seconds
horizontal distance the ball drop in that time
H= (0)(0.4)+1/2(-9.8)(0.4)2
H= -0.78m
negative sign shows that the height of the ball at the net from the top.
Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m
As 1.62m>0.9m so the ball will clear the net.
H_1= V0y t’ + ½ g t’^2
-2.4= (0)t’ + ½ (-9.8) t’^2
t’= 0.69s
X’=V0x t’
X’=(30)(0.96)
X’= 20.7m
You would flip forward or to the side