Final velocity is equal to initial velocity plus at (where a is acceleration and t is time), so Vf = Vi + at
Using that formula;
Vf = 0 + 12.27(3.19)
Vf = 39.14 m/s
Note: you started from rest, so your initial velocity is 0.
Answer:
D) quadruple.
Explanation:
Assuming the same constant acceleration a in both cases, as we have as givens the acceleration a, the distance d, and the initial velocity v, we can use the following kinematic equations in order to compare the distances:
vf² - v₀² = 2*a*d
As the final state of the car is at rest, the final velocity vf, is 0.
⇒ - v₀² = 2*(-a)*d ⇒ d =v₀² / 2*a
1) initial velocity v₀
d₁ = v₀² / 2 a
2 ) initial velocity 2*v₀
⇒ d₂ = (2*v₀)² / 2*a = 4*v₀² / 2*a ⇒ d₂ = 4* (v₀² / 2*a)
⇒ d₂ = 4* d₁
As the equation shows, the distance required to stop, if the initial velocity were doubled, the distance required to stop would quadruple.
Answer:
B. The number of wave cycles that pass through a specific point within a given time period.
Answer:
3) 52.5 m
Explanation:
d = v₁t + 1/2at²
d = 16 m/s (5s) + 1/2(-2.2m/s²) (5s)² = 52.5 m
Answer:
SPST-2
Explanation:
The terms post and pull are also used to describe the contact variations of the switch. The number of "poles" is the number of electrically separated switches that are controlled by a single physical actuator.
For example,
The "SPST" switch is single-pole, single-throw, a simple on / off switch: the two terminals are connected or disconnected from each other.
The "SPDT" switch is single pole, double throw, a simple interrupt changeover switch before making: C (COM, Common) is connected to L1 (NC, normally closed) or L2 (NO, normally open).
"DPST" is bipolar, single shot, equivalent to two SPST switches controlled by one circuit.
"DPDT" is bipolar, double throw, equivalent to two SPDT switches controlled by two circuits.
"3PDT" is 3 pole, double throw, equivalent to 3 SP and DT switches controlled by two circuits.
