Answer:
The air in the soccer ball in cold weather will decrease slightly in size and it becomes flat. The air in the soccer ball in hot weather will seem flat because the low preasure leads to lower bounce in the ball.
The metal door frame in cold weather contracts and the wood contracts more in the winter. The metal door frame in hot weather thermal blowing can occur on the outer surface of the metal door frame. Hopefully that is what you were looking for have a good day.
D. It happens all the time
Answer:
274N 0.41
Explanation:
As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.
then
<em>friction=mass x gravity x sin(21)</em>
Fr=78kg x 9.8m/s2 x sin(21)=274N
<em>friction= coefficient of kinetic friction x normal force of from the slope</em>
Fr= u x 78kg x 9.8m/s2 x cos(21)=274N
Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41
Question:
The options are
a. Technician A only
b. Technician B only
c. Both A and B
d. Neither A nor B
Answer:
The correct option is
d. Neither A nor B
Explanation:
Here, we note that to locate the receiver/driers we look at the high-pressure side of refrigerant system normally around the tubes in between the expansion valve outlet and the condenser outlet and connected to the condenser.
While the accumulator can be found, attached to the evaporator outlet within the low pressure side of the refrigerant system.
Therefore, neither Technician A nor B is correct.
Answer:
P₁ = 2.3506 10⁵ Pa
Explanation:
For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
A₁ v₁ = A₂ v₂
Let's look for the areas
r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm
r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm
A₁ = π r₁²
A₁ = π 1.125²
A₁ = 3,976 cm²
A₂ = π r₂²
A₂ = π 0.1²
A₂ = 0.0452 cm²
Now with the continuity equation we can look for the speed of water inside the hose
v₁ = v₂ A₂ / A₁
v₁ = 11.2 0.0452 / 3.976
v₁ = 0.1273 m / s
Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)
P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂
P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂
Let's calculate
P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25
P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴
P₁ = 2.3506 10⁵ Pa
