Explanation:
Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).
For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:
k is constant of proportionality
It is required to solve the above equation for k
Answer:
a)1.51*10^-22joules b) 1.89*10^-7m
Explanation:
Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules
b) net force acting to stop the proton = 8.01*10^-16
Work done needed to stop the proton = net force acting opposite the motion * distance
Distance covered = need work done/ net force
Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m
Look that one up in you text book PG:678 that is if you got the same book as my friend<span />
Answer:
ididate is a good one and
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J
