The mean may be calculated by summing the values of the refractive index and dividing the sum by the number of experiments. This is:
Mean = (1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53)/6
Mean = 1.51
The mean absolute error is the sum of the absolute values of errors divided by the number of trials:
MAE = (|1.45-1.51|+|1.56-1.51|+|1.54-1.51|+|1.44-1.51|+|1.54-1.51|+|1.53-1.51|)/6
MAE = 0.043
The fractional error is the MAE divided by the actual value:
Fractional error = 0.043 / 1.51
Fractional error = 43/1510
The percentage error is the fractional error multiplied by 100:
Percentage error = 2.85%
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Answer:
5500000 millimeters
Explanation:
1 kilometre= 1000 meter
5.5 km=5.5 * 1000
=5500
Now,
1 metre = 1000 millimetres
5500 metre=1000*5500
=5500000 mm
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
