<span>320. seconds
The ideal gas law is
PV = nRT
where
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = the ideal gas constant
T = absolute temperature of the gas.
Since we're going to want the volume, solve for V
PV = nRT
V = nRT/P
755 mmHg converts to 100658.11 Pascals
25 C = 298.15 K
Let's calculate nT/P, then we'll multiply by R
1 mol * 298.15 K / 100658.11 Pa = 0.002962007 K mol/Pa
The value for R in the most convenient units is 8.3144598 m^3 Pa/(K mol), so
0.002962007 K mol/Pa * 8.3144598 m^3 Pa/(K mol) = 0.024627486 m^3
So 1 mole of air at the specified temperature and pressure has a volume of 24.627 liters. The rest of the problem is now trivial. Just divide by the rate of consumption, so
24.627 l / 0.0770 l/s = 319.8374798 s
Rounding the result to 3 significant figures gives 320. seconds.</span>
Answer:
<h2> 27m/s</h2>
Explanation:
Given data
initital velocity u=15m/s
deceleration a=3m/s^2
time t= 4 seconds
final velocity v= ?
Applying the expression
v=u+at------1
substituting our data into the expression we have
v=15+3*4
v=15+12
v=27m/s
The velocity after 4 seconds is 27m/s
Answer:
f = 0.4 Hz
Explanation:
The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:
f = (1/2π)√(ac/r)
where,
f = frequency of rotation = ?
ac = radial acceleration = 34.1 m/s²
r = radius = length of beam = 5.55 m
Therefore,
f = (1/2π)√[(34.1 m/s²)/(5.55 m)]
<u>f = 0.4 Hz</u>
Answer:The greater the mass, there greater then gravitational pull on their objects.
There greater the distance,the lower the gravitational pull between the object
Explanation:
Gravitational pull(force) is directly proportional to the products Of The masses. therefore if the mass increase,the gravitational pull(force) also increases.
Gravitational pull(force) is inversely proportional to distance.if the distance between the objects increases,the gravitational pull(force) decreases .
