Answer:
P₁ = 1.0 atm
V₁ = 35 L
P₂ = 0.75 atm
Formula:
P₁V₁ = P₂V₂
Solution:
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = (1.0 atm)(35 L) / 0.75 atm
V₂ = 47 L
Final Answer:
V₂ = 47 L
The box is in equilibrium, so Newton's second law says
<em>n</em> + (-<em>w</em>) = 0
65 N + (-<em>f</em> ) = 0
where <em>n</em> denotes the magnitude of the normal force, <em>w</em> denotes the weight of the box, and <em>f</em> denotes the magnitude of the friction force.
The box has a weight of
<em>w</em> = (25 kg) (9.80 m/s²) = 245 N
so <em>n</em> = 245 N, too.
The friction force has magnitude
<em>f</em> = 65 N
and is proportional to the normal force by a factor of <em>µ</em>, the coefficient of kinetic friction. So we have
65 N = <em>µ</em> (245 N) → <em>µ</em> ≈ 0.26
<h2>
Answer:Protons interact in ways that electrons do not. ... Electrons are not affected by the strong force, and so they only get trapped by the electrical attraction to the nucleus which is much weaker in ionized atoms.</h2><h2 /><h2>
Explanation:Therefore it is easier for electrons to move away from one atom to another, transferring charge.</h2>
Answer:
Equal to 5000N
Explanation:
The stress on the material is defined by force per unit of cross-sectional area. So it depends on the force and the diameter of the wire, which is the same for both wires. The material that defines the breaking point, is also the same. Therefore, both wires have their breaking point the same at 5000N. The wire length plays no role in here.
Answer:
a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C
, c) U = 3.21 10⁻¹¹ J
Explanation:
a) The capacitance of a capacitor is
C = k e₀ A / d
Let's calculate
C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²
C = 4,012 10⁻¹⁴ F
b) let's look the charge
C = Q / ΔV
Q = C ΔV
Q = 4,012 10⁻¹⁴ 400
Q = 1.6 10⁻¹¹ C
c) The stored energy
U = ½ C ΔV²
U = ½ 4,012 10⁻¹⁴ 400²
U = 3.21 10⁻¹¹ J
