Answer:
v = 5.88 10⁷ m / s
Explanation:
For this exercise we use the relation
E = m c²
also indicate that all energy is converted into kinetic energy
E = K = ½ (M-2m) v²
where m is the mass of antimatter and M is the mass of the ship's mass. Factor two is due to the fact that equal amounts of matter and antimatter must be combined
we substitute
m c² = ½ (M-2m) v²
v² =
let's calculate
v =
v =
v = 5.88 10⁷ m / s
Answer:
q=1.7346×10⁻⁶C
Explanation:
Since the electric field is perpendicular to the bottom and top of the cube,the total flux is equals the flux over the top of surface plus the flex over the lower surface
Ф(total)=Ф₃₀₀+Ф₂₃₀
But the flux is given by Ф=E.A=EACos(θ) where θ is the angle between Area vector and electric field
So
Ф(total)=E₃₀₀A Cos(180)+E₂₃₀ACos(0)
Ф(total)=A(E₃₀₀ - E₂₃₀)
The total flux is given by Gauss Law as:
Ф(total)=q/ε₀
q=ε₀Ф(total)
q=ε₀(A(E₃₀₀ - E₂₃₀))
Substitute the given values
q=(8.85×10⁻¹²){(70²)(100 - 60)}
q=1.7346×10⁻⁶C
Answer:
0 m/s²
Explanation:
Draw a free body diagram (see attached).
Sum of the forces normal to the incline:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
Sum of the forces parallel to the incline:
∑F = ma
mg sin θ − f = ma
mg sin θ − Nμ = ma
Substituting:
mg sin θ − (mg cos θ) μ = ma
g sin θ − g μ cos θ = a
a = g (sin θ − μ cos θ)
Given that g = 9.8 m/s², θ = 30°, and μ = 1/√3:
a = (9.8 m/s²) (sin 30° − 1/√3 cos 30°)
a = (9.8 m/s²) (1/2 − 1/2)
a = 0 m/s²
The acceleration down the incline is 0 m/s².
The answer is D. time really does pass more slowly in a rest frame of reference relative to a frame of reference that is moving