Given:
ΔT = 38 - 26 = 12°C, temperature change
Q = 11.3 J, heat input
c = 0.128 J/(g-°C), specific heat of lead
Let m = the mass of the lead.
Then
Q = m*c*ΔT
(m g)*(0.128 J/(g-°C))*(12 °C) = 11.3 J
1.536m = 11.3
m = 7.357 g
Answer: 7.36 g (2 sig. figs)
Answer: C. gravitational kinetic
Explanation: Gravitational potential energy is the energy calculated from an object's mass height and the acceleration due to gravity. The gravitational potential energy is the energy an object has as a result of the position of the object in a gravitational field.
Answer: q2 = -0.05286
Explanation:
Given that
Charge q1 = - 0.00325C
Electric force F = 48900N
The electric field strength experienced by the charge will be force per unit charge. That is
E = F/q
Substitute F and q into the formula
E = 48900/0.00325
E = 15046153.85 N/C
The value of the repelled second charge will be achieved by using the formula
E = kq/d^2
Where the value of constant
k = 8.99×10^9Nm^2/C^2
d = 5.62m
Substitutes E, d and k into the formula
15046153.85 = 8.99×10^9q/5.62^2
15046153.85 = 284634186.5q
Make q the subject of formula
q2 = 15046153.85/ 28463416.5
q2 = 0.05286
Since they repelled each other, q2 will be negative. Therefore,
q2 = -0.05286
