Answer:
592.92 x 10³ Pa
Explanation:
Mole of ammonia required = 10 g / 17 =0 .588 moles
We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .
From the relation
PV = nRT
P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )
P = 592.92 x 10³ Pa
Answer:
0.456033049
Explanation:
where N=mg hence
where m is mass of object, g is acceleration due to gravity whose value is taken as 
, 
is the coefficient of static friction and F is the applied force.
Making the subject we obtain
and substituting m for 38 Kg, g for and 170 N for F we obtain
Therefore, the coefficient of static friction is 0.456033049
Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]
From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)
(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)
(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
