Answer:
v₃ = 1.625 [m/s]
Explanation:
To solve this problem we must use the definition of linear momentum conservation, which tells us that momentum is conserved before and after a collision.
Since the collision is inelastic, the two bodies are joined after the collision.
P = m*v [kg*m/s]
m = mass [kg]
v = velocity [m/s]
where:
P = lineal momentum [kg*m/s]
Now, it is important to clarify that in the following equation we will take the left side of the equation as the momentum before the collision and the right side of the equal sign as the momentum after the collision.
Pbefore = Pafter
(m₁*v₁) + (m₂*v₂) = (m₁+m₂)*v₃
where:
m₁ = mass one = 5 [kg]
v₁ = velocity of the mass one = 2 [m/s]
m₂ = mass two = 3 [kg]
v₂ = velocity of the mass two = 1 [m/s]
v₃ = velocity of the combined masses after the collision [m/s]
Now replacing we have:
(5*2) + (3*1) = (5 + 3)*v₃
10 + 3 = 8*v₃
v₃ = 13/8
v₃ = 1.625 [m/s]
Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
Answer:
Hard water is formed when water percolates through deposits of limestone, chalk or gypsum which are largely made up of calcium and magnesium carbonates, bicarbonates and sulfates.
Explanation:
Hard water is formed when water percolates through deposits of limestone, chalk or gypsum which are largely made up of calcium and magnesium carbonates, bicarbonates and sulfates.
Answer:
a.![6.76 m/s^2](https://tex.z-dn.net/?f=6.76%20m%2Fs%5E2)
b.30.4 N
Explanation:
We are given that
Mass of block=m=1.1 kg
Mass of cart=M=3.4 kg
Total mass=m+M=1.1+3.4=4.5 Kg
![\mu=0.69](https://tex.z-dn.net/?f=%5Cmu%3D0.69)
1.We have to find the maximum acceleration such that block does not slide on the cart.
Normal force=![\mu mg](https://tex.z-dn.net/?f=%5Cmu%20mg)
![\mu mg=ma](https://tex.z-dn.net/?f=%5Cmu%20mg%3Dma)
![a=\mu g](https://tex.z-dn.net/?f=a%3D%5Cmu%20g)
Substitute the values
![a=0.69\times 9.8=6.76 m/s^2](https://tex.z-dn.net/?f=a%3D0.69%5Ctimes%209.8%3D6.76%20m%2Fs%5E2)
Using ![g=9.8m/s^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fs%5E2)
Hence, the maximum acceleration such that block does not slide on the cart=![6.76 m/s^2](https://tex.z-dn.net/?f=6.76%20m%2Fs%5E2)
b.![F=ma_{max}](https://tex.z-dn.net/?f=F%3Dma_%7Bmax%7D)
Using the formula
![F=4.5\times (6.76)=30.4 N](https://tex.z-dn.net/?f=F%3D4.5%5Ctimes%20%286.76%29%3D30.4%20N)
Hence, the magnitude of the force on the cart required to provide the maximum acceleration=30.4 N
It’s 1.255 i believe ! good luck ahah