The answer would be C. It will decrease with descent. Hope this helps!
Answer: The correct option is C(both A and B)
Explanation:
In engine chambers, internal combustion of fuel is used to produce mechanical energy.
RTV ( Room Temperature Vulcanising) sealants are silicon products used for the sealing purpose of the engine component. As the name implies once exposed to air, it solidifies. However, it has a negative effect on the oxygen sensor of the engine components by makes a coating over the sensors with the silicone film and, after a time, will affect the operation performed by them. Therefore Technician A is correct.
Technician B is equally correct in that an engine need to be lubricated with oil to help cool the moving part while on. But in order to ensure that oil reaches all movable parts a common method know as heavy-duty reversible spin drills is applied. This method regulates the flow of oil using the engine.
Answer:
r = 4.44 m
Explanation:
For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid
B = ρ g V
Now let's use Newton's equilibrium relationship
B - W = 0
B = W
The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)
σ = W / A
W = σ A
The area of a sphere is
A = 4π r²
W = W₁ + σ 4π r²
The volume of a sphere is
V = 4/3 π r³
Let's replace
ρ g 4/3 π r³ = W₁ + σ 4π r²
If we use the ideal gas equation
P V = n RT
P = ρ RT
ρ = P / RT
P / RT g 4/3 π r³ - σ 4 π r² = W₁
r² 4π (P/3RT r - σ) = W₁
Let's replace the values
r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000
r² (11.81 r -0.060) = 13000 / 4pi
r² (11.81 r - 0.060) = 1034.51
As the independent term is very small we can despise it, to find the solution
r = 4.44 m
Answer:
a. N = 2.49W b. 0.40
Explanation:
a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?
Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s
Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m
The rider's weight W = mg = 9.8m
The ratio of the normal force to the rider's weight is
N/W = 24.43m/9.8m = 2.49
So the normal force expressed in term's of the rider's weight is
N = 2.49W
b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?
The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.
Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.
So, the normal force equals
N = F/μ = W/μ = mg/μ = mrω²
μ = mg/mrω²
= W/N
= 9.8m/24.43m
= 0.40
Answer:
1,520.00 calories
Explanation:
Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).
Given that ĉ = Q / (m.ΔT),
where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),
m= mass of the substance (unity: grams or g), and
ΔT= difference of temperature (unity: Celsius degrees or °C); and
m= 95g and ΔT= 16°C:
Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal
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