Answer:
Estimate money spent on the appliance( 9 Watt bulb ) on a monthly basis is $0.7776
Explanation:
Given the data in the question;
we are to pick a small appliance in our home that we use regularly and look up on the internet how much power it operates at;
so lets consider a 9 Watt bulb
Power p = 9 Watt
Household Voltage = 120 v
we know that; Power = Current I × Voltage V
so,
9 = I × 120
I = 9 / 120
I = 0.075 A
Resistance;
R = V / I
R = 120 / 0.075
R = 1600 Ω
Now, Energy consumed in 1 month(30 days) will be;
E = P × Τ
E = ( 9 / 1000kW ) × ( 30 × 24hrs )
E = 0.009 × 720 hrs
E = 6.48 kWh
Now, the cost for one month will be;
Cost = 6.48 × $0.12
Cost = $0.7776
Therefore, estimate money spent on the appliance( 9 Watt bulb ) on a monthly basis is $0.7776
Answer:
2.726472 s more or 1.5874 times more time is taken than 10-lb roast.
Explanation:
Given:
- The cooking time t is related the mass of food m by:
t = m^(2/3)
- Mass of roast 1 m_1 = 20 lb
- Mass of roast 2 m_2 = 10 lb
Find:
how much longer does a 20-lb roast take than a 10-lb roast?
Solution:
- Compute the times for individual roasts using the given relation:
t_1 = (20)^(2/3) = 7.36806 s
t_2 = (10)^(2/3) = 4.641588 s
- Now take a ration of t_1 to t_2, to see how many times more time is taken by massive roast:
t_1 / t_2 = (20 / 10)^(2/3)
- Compute: t_1 / t_2 = 2^(2/3) = 1.5874 s
- Hence, a 20-lb roast takes 1.5874 times more seconds than 10- lb roast.
t_2 - t_1 = 2.726472 s more
Answer:
a). la acelaración del automóvil es
b). la distancia recorrida por el automóvil es
La aceleración puede ser encontrada por medio de la ecuación cinemática que corresponde a un movimiento rectilíneo uniformemente variado.
(1)
Donde es la velocidad final y es la velocidad inicial.
La velocidad inicial del automóvil sera cero () ya que parte del reposo.
Antes de remplazar los valores en la ecuación 1 es necesario expresar el tiempo en unidades de segundos.
⇒
La rapidez final puede ser expresada en unidades de :
Por lo tanto, la acelaración del automóvil es
Luego para calcular la distancia se puede utilizar la siguiente ecuación.
(2)
Por lo tanto, la distancia recorrida por el automóvil es