We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
</span>F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2<span>μ Fn/m</span> = Δd. This is equal to
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>
Answer:
the train is moving at the speed of v = 1.79 m/s
Explanation:
given,
rain drop is falling vertically down with the speed of = 3.84 m/s
angle of the rain drop = 25°
tan θ = 
tan 25° = 
v =3.84 × tan 25°
v = 1.79 m/s
hence, the train is moving at the speed of v = 1.79 m/s
Answer:
All of these answers are dependent upon the specific scenario, but here are some general answers.
1. An object with a greater height will have more potential energy.
2. Potential energy can be changed into kinetic energy as an object falls. It loses height (potential energy) and gains speed (kinetic energy).
3. Depends on what scenario your class had.
Explanation:
a hot toothbrush. ......lol
