Answer:
18750 kg-m/s
Explanation:
Momentum = mass x velocity
Answer:
E_particle = 1,129 10⁻²⁰ J / particle
T= 817.5 K
Explanation:
Energy is a scalar quantity so it is additive, let's look for the total energy of each gas
Gas a
E_a = 2 5000 = 10000 J
Gas b
E_b = 3 8000 = 24000 J
When the total system energy is mixed it is
E_total = E_a + E_b
E_total = 10000 + 24000 = 34000
The total mass is
M = m_a + m_b
M = 2 +3 = 5
The average energy among the entire mass is
E_averge = E_total / M
E_averago = 34000/5
E_average = 6800 J
One mole of matter has Avogadro's number of atoms 6,022 10²³ particles
Therefore, each particle has an energy of
E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³
E_particle = 1,129 10⁻²⁰ J / particle
For find the temperature let's use equation
E = kT
T = E / k
T = 1,129 10⁻²⁰ / 1,381 10⁻²³
T = 8.175 102 K
T= 817.5 K
Answer: The forces acting on both of them will increase in magnitude.
Explanation:
According to Coulomb's law, the electrostatic force between two bodies is proportional to the product of their two charges. If the charge on A is increased this product increases in size (it must have been non-zero to begin with, since there was a force between them at first). Thus, the force between them rises.
Answer:
1. a ![W_t=746.63 J](https://tex.z-dn.net/?f=W_t%3D746.63%20J)
b ![E_p=212.415 J](https://tex.z-dn.net/?f=E_p%3D212.415%20J)
c ![W_n=183.96J](https://tex.z-dn.net/?f=W_n%3D183.96J)
2. ![T_e=99.71J](https://tex.z-dn.net/?f=T_e%3D99.71J)
Explanation:
a). The work done by the tension is:
![W_t=T*dt](https://tex.z-dn.net/?f=W_t%3DT%2Adt)
![dt=\frac{5.1}{cos(17)}](https://tex.z-dn.net/?f=dt%3D%5Cfrac%7B5.1%7D%7Bcos%2817%29%7D)
![W_t=140N*\frac{5.1}{cos(17)}](https://tex.z-dn.net/?f=W_t%3D140N%2A%5Cfrac%7B5.1%7D%7Bcos%2817%29%7D)
![W_t=746.63 J](https://tex.z-dn.net/?f=W_t%3D746.63%20J)
b). The work done potential of gravity
![E_p=m*g*h](https://tex.z-dn.net/?f=E_p%3Dm%2Ag%2Ah)
![h=5.1*sin(17)](https://tex.z-dn.net/?f=h%3D5.1%2Asin%2817%29)
![E_p=8.5kh*9.8*5.1*sin(30)](https://tex.z-dn.net/?f=E_p%3D8.5kh%2A9.8%2A5.1%2Asin%2830%29)
![E_p=212.415 J](https://tex.z-dn.net/?f=E_p%3D212.415%20J)
c). The work done by the normal force
![W_n=N*d_n](https://tex.z-dn.net/?f=W_n%3DN%2Ad_n)
![W_n=8.5kg*9.8*cos(30)*2.55](https://tex.z-dn.net/?f=W_n%3D8.5kg%2A9.8%2Acos%2830%29%2A2.55)
![W_n=183.96J](https://tex.z-dn.net/?f=W_n%3D183.96J)
2. The increase in thermal energy is:
![T_e=F*d](https://tex.z-dn.net/?f=T_e%3DF%2Ad)
![F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)](https://tex.z-dn.net/?f=F_k%3Du_k%2Am%2Ag%3D0.271%2A8.5kg%2A9.8%2Acos%2830%29)
![F_k=19.5N](https://tex.z-dn.net/?f=F_k%3D19.5N)
![T_e=19.55*5.1m](https://tex.z-dn.net/?f=T_e%3D19.55%2A5.1m)
![T_e=99.71J](https://tex.z-dn.net/?f=T_e%3D99.71J)
mass of the bottle in each case is M = 0.250 kg
now as per given speeds we can use the formula of kinetic energy to find it
1) when speed is 2 m/s
kinetic energy is given as
![K = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![K = \frac{1}{2}(0.250)(2)^2 = 0.5 J](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.250%29%282%29%5E2%20%3D%200.5%20J)
2) when speed is 3 m/s
kinetic energy is given as
![K = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![K = \frac{1}{2}(0.250)(3)^2 = 1.125 J](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.250%29%283%29%5E2%20%3D%201.125%20J)
3) when speed is 4 m/s
kinetic energy is given as
![K = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![K = \frac{1}{2}(0.250)(4)^2 = 2 J](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.250%29%284%29%5E2%20%3D%202%20J)
4) when speed is 5 m/s
kinetic energy is given as
![K = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![K = \frac{1}{2}(0.250)(5)^2 = 3.125 J](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.250%29%285%29%5E2%20%3D%203.125%20J)
5) when speed is 6 m/s
kinetic energy is given as
![K = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![K = \frac{1}{2}(0.250)(6)^2 = 4.5 J](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.250%29%286%29%5E2%20%3D%204.5%20J)