A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?
<span>Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. </span>
<span>At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. </span>
<span>So total time is 8.5 + 25.5 or 34 sec. </span>
<span>Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.</span>
I Think The answer is d I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you
Is that the full question?
Respuesta:
2 × 10⁴ V
Explicación:
Paso 1: Información provista
- Carga transportada (q): 4 nC
- Trabajo realizado (W): 7 × 10⁻⁵ J
Paso 2: Convertir q a Coulomb
Usaremos el factor de conversión 1 C = 10⁹ nC.
4 nC × 1 C/10⁹ nC = 4 × 10⁻⁹ C
Paso 3: Calcular el potencial eléctrico (V) de la esfera
Usaremos la siguiente fórmula.
V = W/q
V = 7 × 10⁻⁵ J/4 × 10⁻⁹ C = 2 × 10⁴ V
