The statement would be False. T<span>he potential energy of a membrane potential comes solely from the difference in electrical charge across the membrane. In addition to that, membrane potential actually regulates the potential difference of nerve cells across the membrane estimated at 70 mV.</span>
C. It is answered by observation and evidence.
Good scientific explanations are defined, measurable and controllable. They can be answered by an experiment.
Answer:
L = 1.11 x
m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Explanation:
Solution:
Data Given:
Heat Energy = 52000 J
Dielectric Constant of the plastic Bag = 3.7 = K
Thickness = 2.6 x
m =d
V = 610 volts
A = width x Length
width = 20 cm = 20 x
m
Length = ?
So,
we know that,
U = 1/2 C Δ
U = 52000 J
C = ?
V = 610 volts'
So,
U = 1/2 C Δ
52000 J = (0.5) x (C) x (
)
C = 0.28 F
And we also know that,
C = 
E = 8.85 x 
K = 3.7
A = 0.20 x L
d = 2.6 x
m
Plugging in the values into the formula, we get:
0.28 = 
Solving for L, we get:
L = 1.11 x
m,
is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Answer:
10.4 m/s
Explanation:
First, find the time it takes for the projectile to fall 6 m.
Given:
y₀ = 6 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.11 s
Now find the horizontal position of the target after that time:
Given:
x₀ = 6 m
v₀ = 5 m/s
a = 0 m/s²
t = 1.11 s
Find: x
x = x₀ + v₀ t + ½ at²
x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²
x = 11.5 m
Finally, find the launch velocity needed to travel that distance in that time.
Given:
x₀ = 0 m
x = 11.5 m
t = 1.11 s
a = 0 m/s²
Find: v₀
(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²
v₀ = 10.4 m/s