Answer:
3727.24km/h
Explanation:
Hello!
To solve this problem we must first know what is the distance from the earth to the moon, this will be our radius.
=384400Km
Then we find the distance traveled which would be the perimeter of a circle = 2πr, finally to find the speed we divide the distance traveled by 27 days.
Finally we use conversion factor to have the speed in km / h
solving
the moon is orbiting at speed of 3727.24km/h
<span>We can draw a free body diagram for the safe to show all the horizontal forces acting on the safe. The three horizontal forces are Clyde's force, Bonnie's force, and friction. Since the safe slides at a constant speed (no acceleration), the net force must be zero.
0 = Clyde's force + Bonnie's force - friction
friction = Clyde's force + Bonnie's force
mg mu = Clyde's force + Bonnie's force
mu = (Clyde's force + Bonnie's force) / (mg)
mu = (445 N + 350 N) / (300 kg x 9.80~m/s^2)
mu = 0.27
The coefficient of friction is 0.27</span>
Answer:
7.35 J
Im assuming, upon answering the question, that the gravity in this scenario is 9.8? As 9.8 is the gravitational force upon the earth.
Answer:
E = α/2∈₀ [ 1 - a²/r² ]
Ф = α/2∈₀
Explanation:
Using Gauss Law:
ρ(r) = a/r, dA
= 4 π r²d r
Ф = ρ(r')dA
Ф = ρ(r')dA
= 4πα r'dr'
Ф = 4 π α 1/2(r²-a²)
E(4πr²) = ∈₀
= ∈₀(4πr²)
= α (r² - a²) / 2 ∈₀ (r²)
= α/2∈₀ [ r²/r² - a²/r² ]
E = α/2∈₀ [ 1 - a²/r² ]
Electric field of the point charge:
E = q / 4π∈₀r²
= α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²
For to be constant:
- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²
-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀
= α - αa² + αa² / 2 ∈₀
= α /2 ∈₀
Hence:
Ф = α/2∈₀