Washing your hands and taking medicine
Answer:
Lines of credit are unsecured loans. That means the bank is taking a huge risk. The bank has to be certain the borrower has a credit history that indicates (s)he will pay back the loan
"Electrostatic forces are attractive or repulsive forces between particles that are caused by their electric charges."
Answer:
a)
two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)
b)
both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
Explanation:
Given that;
L = 0.26 m
k = 180 N/m
x = 0.039 m
a)
we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.
b)
Spring force F = kx
F = 180 × 0.039
F = 7.02 N
Now, Electrostatic force F = Keq²/r²
where r = L + x = ( 0.26 + 0.039 )
we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2
so from the equation; F = Keq²/r²
Fr² = Keq²
q = √ ( Fr² / Ke )
we substitute
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ (0.627595 / 9×10⁹)
q = √(6.97 × 10⁻¹¹)
q = 8.35 × 10⁻⁶ C
Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
