Answer:
The temperature of the gas is 1197.02 K
Explanation:
From ideal gas law;
PV = nRT
Where;
P is the pressure of the gas
V is the volume of the gas
R is ideal gas constant = 8.314 L.kPa/mol.K
T is the temperature of the gas
n is the number of moles of gas
Volume of the gas in the cylindrical container = πr²h
Given;
r = 6/2 = 3 cm = 0.03 m
h = 11 cm = 0.11 m
V = π × (0.03)² × 0.11 = 3.11 × 10⁻⁴ m³ = 0.311 L
number of moles of oxygen gas = Reacting mass / molar mass
Therefore, the temperature of the gas is 1197.02 K
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
Answer:
Explanation:
delta V = v * alpha * delta T
= V * 0.00053 * (92.2 - 55.0)
= 0.019716 V
percentage that the owner
= [delta V / V] * 100
= [0.019716 V / V] * 100
= 1.9716 %
First of all, we can find the mass of the person, since we know his weight W:
And so
We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
We know the acceleration of the system,
(upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
