Answer:
Explanation:
We'll use the momentum equation:
where:
p = momentum
m = mass
v = velocity
Since we're doing the magnitude of momentum of the system, we'll add the mass of the cyclist and the mountain bike together:
Given that, we can now substitute our given values into the momentum equation:
Our final answer is:
Answer: 0.09 m
Explanation:
Centripetal acceleration in terms of tangential speed is:
where r is the radius.
It is given that,
centripetal acceleration of the mouse,
tangential speed ,
Radius of salami is:
Thus, the radius of the salami is 0.09 m.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
= K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
Answer:
a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing
b) E_total = 1.89 10⁶ N / C, field is incoming radial
c) E_total = 0
Explanetion:
For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is
Ф = E .dA = q_{int} /ε₀
inside the spherical shell there are no charges
The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.
We have two sphere shells with radii 0.050m and 0.15m respectively
a) point where you want to know the electric field d = 0.20 m
shell 1
the point is on the outside,d>ro, therefore we can consider the charge to be concentrated in the center
E₁ = k q₁ / d²
shell 2
the point is on the outside,d>ro
E₂ = k q₂ / d²
the total camp is
E_total = -E₁ + E₂
E_total = k ( )
E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²
E_total = 6,525 10⁵ N /C
The field direction is radial and outgoing ti the shells
b) the calculation point is d = 0.10m
shell 1
point outside the shell d> ro
E₁ = k q₁ / d²
shell 2
the point is inside the shell d <ro
Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero
E₂ = 0
E_total = E₁
E_total = k q₁ / d²
E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²
E_total = 1.89 10⁶ N / A
As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1
c) the point of interest d = 0.025 m
shell 1
point is inside the shell d< ro
as there are no charges inside
E₁ = 0
shell 2
point is inside the radius of the shell d <ro
E₂ = 0
the total field is
E_total = 0