Answer:
230.4 N
Explanation:
From the question given above, the following data were obtained:
Charge (q) of each protons = 1.6×10¯¹⁹ C
Distance apart (r) = 1×10¯¹⁵ m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The force exerted can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²
F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰
F = 2.304×10¯²⁸ / 1×10¯³⁰
F = 230.4 N
Therefore, the force exerted is 230.4 N
Answer:
1.8*10^8m/s
Explanation:
Using
L= lo√1-v²/c²
So making v subject we have
V= c√1-4.8²/6²
V= 0.6*c
V= 0.6*3E8m/s
V= 1.8*10^8m/s
Answer:
232 J/K
Explanation:
The amount of heat gained by the air = the amount of heat lost by the tea.
q_air = -q_tea
q = -mCΔT
q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)
q = 68,000 J
The change in entropy is:
dS = dQ/T
Since the room temperature is constant (isothermal):
ΔS = ΔQ/T
Plug in values (remember to use absolute temperature):
ΔS = (68,000 J) / (293 K)
ΔS = 232 J/K
Force = mass times acceleration
F = 21000 x 36.9 = 774900
Therefore, 774900N force is required.
Answer:
Noble gas element
Explanation:
because Noble gas element can neither loose nor gain electrons