The eccentricity of its orbit is ![$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$](https://tex.z-dn.net/?f=%24%24U%3D-2.13%20%5Ctimes%2010%5E%7B17%7D%20%5Cmathrm%7B~J%7D%20.%5Cend%7Baligned%7D%24%24)
<h3>What is mass?</h3>
- Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
- The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
- For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
- An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.
The length of the semi-major axis is calculated as follows:
where![, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$](https://tex.z-dn.net/?f=%2C%20%24G%3D6.67%20%5Ctimes%2010%5E%7B-1%7D%20%5Cmathrm%7B~m%7D%5E3%20%2F%20%5Cmathrm%7Bkgs%7D%24)
mass of sur
- a mass of the comet
![$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$](https://tex.z-dn.net/?f=%24%24%5Cbegin%7Baligned%7D%5Ctherefore%20%5Cquad%20%5Ctext%20%7B%20At%20aphelion%2C%20%7D%20r%20%26%3D50%20%5Ctimes%20U%20%5C%5C%26%3D50%20%5Ctimes%201.496%20%5Ctimes%2010%5E%7B11%7D%20%5Cmathrm%7B~m%7D%20.%20%5C%5CU%3D-%5Cfrac%7B6.67%20%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%20%5Cmathrm%7Bm%7D%201.99%20%5Ctimes%2010%5E%7B30%7D%20%5Ctimes%201.20%20%5Ctimes%2010%5E%7B10%7D%7D%7B50%20%5Ctimes%201.496%20%5Ctimes%2010%5E%7B11%7D%7D%20%5C%5CU%3D-2.13%20%5Ctimes%2010%5E%7B17%7D%20%5Cmathrm%7B~J%7D%20.%5Cend%7Baligned%7D%24%24)
![$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$](https://tex.z-dn.net/?f=%24%24U%3D-2.13%20%5Ctimes%2010%5E%7B17%7D%20%5Cmathrm%7B~J%7D%20.%5Cend%7Baligned%7D%24%24)
To learn more about mass, refer to:
brainly.com/question/3187640
#SPJ4
(a) 392 N/m
Hook's law states that:
(1)
where
F is the force exerted on the spring
k is the spring constant
is the stretching/compression of the spring
In this problem:
- The force exerted on the spring is equal to the weight of the block attached to the spring:
![F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N](https://tex.z-dn.net/?f=F%3Dmg%3D%282.0%20kg%29%289.8%20m%2Fs%5E2%29%3D19.6%20N)
- The stretching of the spring is
![\Delta x=15 cm-10 cm=5 cm=0.05 m](https://tex.z-dn.net/?f=%5CDelta%20x%3D15%20cm-10%20cm%3D5%20cm%3D0.05%20m)
Solving eq.(1) for k, we find the spring constant:
![k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BF%7D%7B%5CDelta%20x%7D%3D%5Cfrac%7B19.6%20N%7D%7B0.05%20m%7D%3D392%20N%2Fm)
(b) 17.5 cm
If a block of m = 3.0 kg is attached to the spring, the new force applied is
![F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N](https://tex.z-dn.net/?f=F%3Dmg%3D%283.0%20kg%29%289.8%20m%2Fs%5E2%29%3D29.4%20N)
And so, the stretch of the spring is
![\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm](https://tex.z-dn.net/?f=%5CDelta%20x%3D%5Cfrac%7BF%7D%7Bk%7D%3D%5Cfrac%7B29.4%20N%7D%7B392%20N%2Fm%7D%3D0.075%20m%3D7.5%20cm)
And since the initial lenght of the spring is
![x_0 = 10 cm](https://tex.z-dn.net/?f=x_0%20%3D%2010%20cm)
The final length will be
![x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm](https://tex.z-dn.net/?f=x_f%20%3D%20x_0%20%2B%5CDelta%20x%3D10%20cm%2B7.5%20cm%3D17.5%20cm)
The answer would be 2.63. Your welcome. This has been changed to the correct answer.
Answer:
K = m g (A - A2)
Explanation:
In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential
Em = U₀ = m g A
For when the system is at an ele
Elongation A2 less than A, energy has two parts
Em = K + U₂
K = Em –U₂
We substitute
K = m g A - m gA2
K = m g (A - A2)