I'm not sure about the distance to the nearest star, but it's probably about 4 light-years (L-y).
1 L-y = 1.86 * 10E5 mi/sec * 3600 sec/hr * 24 hr/day * 365 day/yr
1 L-y = 5.9 *10E12 mi and 4 L-y = 2.3 *10E13 mi distance to star
2.3 * 10E13 mi / 900 mi/hr = 2.6 * 10E10 hr hours to star
2.6 * 10E10 hr / (24 hr/day) = 1.1 * 10E9 day days to star
1.1 * 10E9 day / 365 day/yr = 3 * 10E6 yr = 3 million years to star
Answer:
Simple, plug 4 into c.
4 + 5 = 13
Is this true, no. So therefore c is not equal to 4.
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
Answer:
65.1 °C
Explanation:
m = mass of the water = 0.5 kg
= initial temperature of water = 35.00 °C
= final temperature of water
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of heat removed from water = 6.3 x 10⁴ J
Amount of heat removed from water is given as
Q = m c ( - )
Inserting the values
6.3 x 10⁴ = (0.5) (4186) ( - 35.00)
= 65.1 °C
Answer:
yes the diagram is correct
Explanation: