Coulomb's law:
Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²
= (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C) / (1.0 m)²
= (8.99×10⁹ x 1×10⁻¹² / 1.0) N
= 8.99×10⁻³ N
= 0.00899 N repelling.
Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.
-- '4.0 kg masses'; don't need it.
Mass has no effect on the electric force between them.
-- 'frictionless table'; don't need it.
Friction has no effect on the force between them,
only on how they move in response to the force.
</span></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.
The bouncy ball experiences the greater momentum change.
To understand why, you need to remember that momentum is actually
a vector quantity ... it has a size AND it has a direction too.
The putty and the ball have the same mass, and you throw them
with the same speed. So, on the way from your hand to the wall,
they both have the same momentum.
Call it " M in the direction toward the wall ".
After they both hit the wall:
-- The putty has zero momentum.
Its momentum changed by an amount of M .
-- The ball has momentum of " M in the direction away from the wall ".
Its momentum changed by an amount of 2M .
Answer:
Explanation:
= Mg ion = 
= F ion = 
q = Charge of electron = 
r = Distance between ions = 
k = Coulomb constant = 
Electrical force is given by
![F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N](https://tex.z-dn.net/?f=F%3D-%5Cdfrac%7Bkq_1q_2%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20F%3D-%5Cdfrac%7B8.99%5Ctimes%2010%5E9%5Ctimes%202%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20-1%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B%5B%280.072%2B0.133%29%5Ctimes%2010%5E%7B-9%7D%5D%5E2%7D%5C%5C%5CRightarrow%20F%3D1.09527%5Ctimes%2010%5E%7B-8%7D%5C%20N)
The attractive force is
This is mr day and infact am seeing this… come into my office after class.
