Einstein's mass energy relation, and the formula for finding the percentage error gives;
7) a) 9.0 × 10¹⁶ Joules
b) The bulb could be powered for approximately 47533146.96 years
8) 16.46%
9) The possible values are;
61.093 Joules and 96.77 Joules
<h3>What is a percentage error?</h3>
Percentage error is the ratio of the difference between estimated value and the actual value to the actual value expressed as a percentage.
7) a) The Einstein mass energy relation is presented as follows;
E = m•c²
The speed of light, <em>c </em>= 3×10⁸ m/s
The mass, <em>m </em>= 1 kg, gives;
E = 1 kg × (3×10⁸ m/s)² = 9.0 × 10¹⁶ Joules
- The energy present in 1 kg. mass is; 9.0 × 10¹⁶ Joules
b) Energy consumption of the light bulb = 60 J/sec
The duration, <em>t</em>, the light bulb will remain lit is therefore;
- t = (9.0 × 10¹⁶)/(60) = 1.5 × 10¹⁵ s ≈ 47533146.96 years
8) Percentage error, %error, is found as follows;
%error = ((Approximate value - Actual value)/(Actual value)) × 100
Appreciate value = 46,620 km
Actual value = 40,030.2 km
Which gives;
%error = ((46,620 - 40,030.2)/(40,030.2))×100 ≈ 16.46%
- The percentage error of Eratosthenes estimate of the circumference of the Earth is approximately 16.46%
9) The calculated value of the kinetic energy = 74.9 J
The observed error in the calculation = 22.6%
Required; The two possibilities for the correct kinetic energy
Solution;
E% = ((Approximate value - Actual value)/(Actual value)) × 100
Taking the calculated value as the approximate value, and the correct value as the actual value, we have;
22.6% = ((74.9 - Actual value)/(Actual value)) × 100
(22.6/100)×Actual value = 74.9 - Actual value
Actual value + 0.226×Actual value = 74.9
1.226 × Actual value = 74.9
Which gives;
- Actual value = 74.9 J/1.226 ≈ 61.093 J
The actual value can also be given as follows;
22.6% = ((Actual value - 74.9)/(Actual value)) × 100
Actual value - 0.226×Actual value = ,74.9
Actual value = 74.9 J/(1-0.226) ≈ 96.77 J
- The possible values for the kinetic energy are approximately; 61.093 J and 96.77 J
Learn more about calculating percentage error in measurements here;
brainly.com/question/5493941
#SPJ1
