Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:
Where,
P = Pressure
V = Volume
T = Temperature
n = amount of substance
R = Ideal gas constant
We start by calculating the volume of inhaled O_2 for it:
Our values are given as
P = 1atm
T=293K 
Using the equation to find n, we have:
Number of molecules would be found through Avogadro number, then
The answer is: [C]: "4" .
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Note: To balance this equation, the coefficient, "4", should be placed in front of the PCl₃ ; and the coefficient, "6", should be placed in front of the Cl₂ .
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The balanced equation is:
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P₄ (s) + 6 Cl₂ (g) <span>→ 4 </span>PCl₃ (l) .
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Answer:
v = 2.94 m/s
Explanation:
When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.
Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.
Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means
(1/2)kx^2 = (1/2)mv^2
kx^2 = mv^2
v^2 = (kx^2)/m
v = sqrt((kx^2)/m)
v = x * sqrt(k/m)
v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg
v = 2.94 m/s
Answer:
The acceleration of the car, a = -3.75 m/s²
Explanation:
Given data,
The initial velocity of the airplane, u = 75 m/s
The final velocity of the plane, v = 0 m/s
The time period of motion, t = 20 s
Using the I equations of motion
v = u + at
a = (v - u) / t
= (0 - 75) / 20
= -3.75 m/s²
The negative sign indicates that the plane is decelerating
Hence, the acceleration of the car, a = -3.75 m/s²
