Answer:
(a)= 264mmHg
(b)= 2000mmHg
(c)474.82mmHg
(d)= 511.63mmHg
Explanation:
the question deals with boyles law, which states that the volume of a given mass of gas at constant temperature is inversely proportional to its pressure
V ∝ 1/P
P₁V₁ = P₂V₂
making V₂ as the subject of formular
P₂ = P₁V₁/ V₂
with a volume of 25.0L
P₂ = 660×10 / 25
= 264mmHg
with a volume of 3.30 L
P₂ = 660 × 10 / 3.30
= 2000mmHg
with a volume of 13900 mL
= 13.9L
P₂ =660× 10 / 13.9
474.82mmHg
with a volume of 12900 mL
P₂ =660×10 / 12.9
= 511.63mmHg
Answer:

Explanation:
given,
density of blood = 1060 kg/m³
v₂ = 0.800 m/s
v₁ = 0.475 m/s
change in pressure calculation
using Bernoulli's equation

both are at same level



the change in pressure is equal to 219.62 Pa
The tension force being supplied by the rope is 245 N.
<h3>What is tension force?</h3>
- Tension force is the force exerted on a rope or cord due to the weight of an object suspended from it.
The tension force on the given rope due to the weight of the lamp hanging from the rope is calculated by applying Newton's second law of motion as shown below;
T = mg
where;
- m is the mass = 25 kg
- g is acceleration due to gravity = 9.8 m/s²
T = 25 x 9.8
T = 245 N
Thus, the tension force being supplied by the rope is 245 N.
Learn more about tension force here: brainly.com/question/12797227
Answer: Distance= 100,000 km
Mass= 15 million kg Mass= 5 million kg
Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K