Answer:
83.33 C
Explanation:
T1 = 111 C, m1 = 2m
T2 = 28 C, m2 = m
c = 0.387 J/gK
Let the final temperature inside the calorimeter of T.
Use the principle of calorimetery
heat lost by hot body = heat gained by cold body
m1 x c x (T1 - T) = m2 x c x (T - T2)
2m x c X (111 - T) = m x c x (T - 28)
2 (111 - T) = (T - 28)
222 - 2T = T - 28
3T = 250
T = 83.33 C
Thus, the final temperature inside calorimeter is 83.33 C.
The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
the 2 main categories of nonmetals are REACTIVE NONMETALS an NOBLE GAS.
hope it helps
Answer:
Negative intrapleural pressure is the correct answer
Explanation:
Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.
Air moves from a region of higher pressure to one of lower pressure. Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.
Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.
Answer:
12 ml
Explanation:
The initial volume in the cylinder is 20 ml
adding the rock adds volume to the cylinder
the new volume is 32 ml .....the increase in volume is the volume of the rock : 32 - 20 = 12 ml volume of rock
