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:
1.312 x 10⁻¹² J/nucleon
Explanation:
mass of ¹³⁶Ba = 135.905 amu
¹³⁶Ba contain 56 proton and 80 neutron
mass of proton = 1.00728 amu
mass of neutron = 1.00867 amu
mass of ¹³⁶Ba = 56 x 1.00728 amu + 80 x 1.00867 amu
= 137.10128 amu
mass defect = 137.10128 - 135.905
= 1.19628 amu
mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg
= 1.9858 x 10⁻²⁷ Kg
speed of light = 3 x 10⁸ m/s
binding energy,
E = mass defect x c²
E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²
E = 17.87 x 10⁻¹¹ J/atom
now,
binding energy per nucleon =
= 0.1312 x 10⁻¹¹ J/nucleon
= 1.312 x 10⁻¹² J/nucleon
Answer:
The work required to stretch a spring 12 ft beyond its natural length is 432 ft-lb
Explanation:
The work to stretch a spring is calculated using the formula:
Equation (1)
W = work in ft-lb
k = spring constant in lb/ft
x = spring deformation in ft
we clear k from the equation (1)
Equation (2)
We replace x = 2ft, W = 12 ft-lb in the equation (2)
Calculation of work required to stretch spring 12 ft
We replace k = 6 lb/ft and x = 12ft in the equation (1)
If a main sequence star is cooling and expanding, it is entering the red giant stage. This means that the star has burned up all of it's hydrogen and is now starting to burn its helium making it cooler.