Answer:
ρ_body = 1000 kg / m³
Explanation:
This is an exercise in fluid mechanics, specifically we must use the Archimedean principle, which states that the thrust is equal to the weight of the dislodged liquid.
In this case let's start by finding the volume of our body
oak block
v = l to h
v = 0.02 0.02 0.05
V = 2 10⁻⁵ m³
cooper block indicate that it has the same dimensions so its volume is the same, the total volume of the body is
V_total = 4 10⁻⁵ m³
as they indicate that the body is fully submerged there is a balance between weight and thrust
B - W = 0
the push is
B = ρ_fluid g V_total
the body weight is
ρ_body = M / V_total
M = ρ_body V_total
W = Mg
W = ρ_body V_total g
we substitute
ρ_fluid g V_total = ρ_body V_total g
ρ_body = ρ_fluid
in this case the body is in equilibrium in the fluid, in case the density of the body is greater than that of the fluid, the body sinks
Therefore the average density is equal to the density of the fluid, since since it is water the density is
ρ_body = 1000 kg / m³
Answer:
U = - 4 x 
Explanation:
ΔV = potential difference =
Volts
q = charge on electron =
C
electric potential energy is given as
U = q ΔV = (
) (
)
= - 4 x 
Refer to the diagram shown below.
d = distance (miles) from Alphaville to Betaville.
v = speed (mph) of the plane with no wind.
With no wind:
The time taken to travel a distance of 2d is 4 hrs, 48 min = 4.8 hrs.
Therefore
2d/v = 4.8
v = 2d/4.8 = 0.4167d mph (1)
With the wind:
The velocity from Alphaville to Betaville is (v + 100) mph.
The time of travel is
t₁ = d/(v+100) h
The velocity from Betaville to Alphaville is (v - 100) mph.
The time of travel is
t₂ = d/)v-100) h
Because the return trip takes 5 hours, therefore
t₁ + t₂ = 5

(2)
From (1), obtain
2(0.4167)d² = 5[(0.4167d)² - 10⁴]
0.8334d² = 0.8682d² - 5 x 10⁴
0.0348d² = 5 x 10⁴
d = 1198.7 mi
Answer: 1199 miles (nearest integer)
Answer:
The two moments must be the same:
p1=p2
m1v1=m2v2
v2=(m1v1)/m2
v2=(90 kg x 0.9 m/s)/110kg=0.7 m/s