Answer:
the buoyant force on the chamber is F = 7000460 N
Explanation:
the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.
Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume
mass of water displaced = density of seawater * volume displaced
m= d * V , V = 4/3π* Rext³
the buoyant force is the weight of this volume of seawater
F = m * g = d * 4/3π* Rext³ * g
replacing values
F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N
Note:
when occupied the tension force on the cable is
T = F buoyant - F weight of chamber = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N
Answer:
4.2 J
Explanation:
Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K
From specific heat capacity,
Q = cmΔt.............................. Equation 1
Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.
Given: m = 1 g = 0.001 kg, Δt = 1 °C
Constant : c = 4200 J/kg.°C
Substitute into equation 1
Q = 0.001×4200(1)
Q = 4.2 J.
Hence the energy absorbed or lost = 4.2 J
Answer:
The wagon will move to the right.
Explanation:
From the question given above, the following data were obtained:
Force applied to the left (Fₗ) = 10 N
Force applied to the right (Fᵣ) = 30 N
Direction of the wagon =.?
To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:
Force applied to the left (Fₗ) = 10 N
Force applied to the right (Fᵣ) = 30 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 30 – 10
Fₙ = 20 N to the right
From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.
Answer:
68.8 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of box = 18 Kg
Coefficient of friction (μ) = 0.39
Force of friction (F) =?
Next, we shall determine the normal force of the box. This is illustrated below:
Mass (m) of object = 18 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal force (N) =?
N = mg
N = 18 × 9.8
N = 176.4 N
Finally, we shall determine the force of friction experienced by the object. This is illustrated below:
Coefficient of friction (μ) = 0.39
Normal force (N) = 176.4 N
Force of friction (F) =?
F = μN
F = 0.39 × 176.4
F = 68.796 ≈ 68.8 N
Thus, the box experience a frictional force of 68.8 N.
In 6 secs, the dog covers-
S=vt
8.9*6 = 53.4 m.
In the same time, the cat covers, 53.4-3.8 = 49.6 m.
Thus, speed of the cat, v= s/t,
= 49.6/6 = 8.267 m/s
