Because she had serious traffic accident on Friday the 13th of last month, Felicia is conceived that all Friday the 13th will br
1 answer:
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Answer:
a) K = 49.5 J b) k = 1378 N / m c) ΔE = 34 J d) μ = 0.399
Explanation:
For this exercise we will use the concepts of energy
a) The initial kinetic energy is
K = ½ m v²
K = ½ 3.96 5²
K = 49.5 J
b) let's use energy conservation
Em₀ = K = ½ m v²
= Ke = ½ k x²
Em₀ =
½ m v² = ½ k x²
k = m v² / x²
k = 3.96 5² / 0.268²
k = 1378 N / m
c) Let's calculate the final energy of the spring
= Ke = ½ k x²
= ½ 1378 0.15²
= 15.5 J
The initial energy is the kinetics of the block
Em₀ = 49.5 J
The lost energy is the difference with the initial
ΔE = - Em₀
ΔE = 15.5 - 49.5
ΔE = - 34 J
the negative sign means that the energy dissipates
d) For this part we use the concept of work
W = F d cos θ = ΔK
In this case the force is the friction force that always opposes displacement, so the angle 180 ° and cos 180 = -1
W = -fr d = ΔK
The force of friction is
fr = μ N
With Newton's second law
N-w = 0
N = W = mg
Let's calculate
-μ mg d = Kf -K₀o
μ = K₀ / mgd
μ = 49.5 / (3.96 9.8 3.20)
μ = 0.399
Answer:
1/9 of that just outside the smaller sphere
Explanation:
The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:
where
k is the Coulomb's constant
Q is the charge on the sphere
r is the distance from the centre of the sphere
Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is
The second sphere has a radius which is 3 times that of the smaller sphere:
So, the electric field just outside the second sphere is
So, the correct answer is 1/9.