In a slide, you’re potential energy (sitting on top of a slide) is transformed into kinetic energy when you slide down and then goes back to potential energy when you reach the bottom at a stop. No energy was destroyed nor created, just transformed from one state to another
Answer:
J.J. Thomson discovered <u><em>the electron</em></u> by experimenting with a Crookes, or cathode ray, tube. He demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas.
Explanation:
the answer is the underlined part of the answer section. I hoped this helped you a lot!! Study hard for whatever you are doing!!
Answer:
a = 30.832 ft/s²
Explanation:
To solve this problem let's start by finding the braking acceleration using kinematics, where the distance is x = 408 ft, the initial velocity vo = 100 mi / h and the final velocity is zero v = 0
v² = v₀² - 2 a x
0 = v₀² - 2ax
a =
Let's start by reducing the magnitudes to ft / s
v₀ = 100 mi / h (5280 foot / 1 mile) (1h / 3600 s) = 146.666 ft / s
let's calculate
a =
a = 26.36 ft / s²
Let's call this acceleration a_effective, this acceleration is in the opposite direction to the speed of the vehicle.
Let's use a rule of three (direct proportions) to find the acceleration applied by the brake system (a1) which has an efficiency of 95%. or 0.95
a₁ =
Let's use another direct proportion rule If the acceleration of the brake system (a₁) for an applied acceleration (a) with an efficiency of 0.90
a =
we substitute
a =
let's calculate
a =
a = 30.832 ft/s²
This is the maximum relationship that the vehicle can have for when it brakes to stop at the given distance
Ok let me help you with this:
<span>
!n case a) Line both up head-to-tail in a straight line... that's the only way to get a sum 2F. In case b) sqrt(2) is the length of the hypotenuse in a 45 degree triangle, so the vectors must be at 90 degrees to each other
in the case of c) lined up head to tail, the only way to get 0 is if they point in opposite direction</span>
Should be able to pull it off the same way you got it on