Formula to find gravitational potential energy:
mgh
m: mass
g: gravitational acceleration
h: height (relative to reference level)
so the P.E. at 1.0.m is (5x9.8x1)= 49J
P.E. at 1.5m is (5x9.8x1.5) =73.5J
P.E. at 2.0m is (5x9.8x2)=98J
<span>(1) </span>Through the Second
Law of motion, the equation for Force is:
F = m x a
Where
m is mass and a is acceleration (deceleration)
<span>(2) </span>Distance is
calculated through the equation,
D
= Vi^2 / 2a
Where
Vi is initial velocity
<span>(3) </span>Work is calculated
through the equation,
W = F x D
Substituting
the known values,
Part
A:
<span>(1) </span> F = (85
kg)(2 m/s^2) = 170 N
<span>(2) </span> D = (37
m/s)^2 / (2)(2 m/s^2) = 9.25 m
<span>(3) </span> W = (170
N)(9.25 m) = 1572.5 J
Part
B:
<span>(1) </span> F = (85 kg)(4
m/s^2) = 340 N
<span>(2) </span>D = (37 m/s)^2 /
(2)(4 m/s^2) = 4.625 m
<span>(3) </span><span> W = (340
N)(4.625 m) = 1572.5 J</span>
Answer:
100 cm³
Explanation:
Use ideal gas law:
PV = nRT
where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.
n and R are constant, so:
P₁V₁/T₁ = P₂V₂/T₂
If we say point 1 is at 40m depth and point 2 is at the surface:
P₂ = 1.013×10⁵ Pa
T₂ = 20°C + 273.15 = 293.15 K
P₁ = ρgh + P₂
P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa
P₁ = 4.933×10⁵ Pa
T₁ = 4.0°C + 273.15 = 277.15 K
V₁ = 20 cm³
Plugging in:
(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)
V₂ = 103 cm³
Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.
As I found out the choices for your question which are:
<span>A) F2 to F-
B) Cr2O7²- → Cr2+
C) O2 to H2O
D) HAsO2 to As
</span>
Unfortunately, the answer does not belong to the choices provided. In fact, it is the oxidation half-reaction that occurs at the anode of an electrode for it to transform chemical energy to consumable electrical energy.
Answer: 0.817A
Explanation:
Assuming , that one coulomb per second of negative charge alone flow through a conductor and no positive charges flow. I.e Q=It
It means a current of one A flow in the opposite direction.
This is similar to one coulomb per second of positive charge flowing through and there is no negative charge,
In addition, the one coulomb per second of positive charge flows. This is flowing in the current direction of the previous one. Then, the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.The charges cannot cancel each other, because even before the current flow the conductor was neutral.
According to this, the current in the given problem is
[2.7 + 2.4] x 10 ^ 18 * 1.602 x 10^ [-19] C/s
= 0.817 A
