A. 2 C₂ H₆ + 7 O₂ → 4 C O₂ + 6 H₂ O
according to law of conservation of mass , the total mass of reactants side must be same as the total mass of product side. so we need to check if each atom in the equation has same number on both side of the equation or not.
in this equation , we have
4 atoms of carbon left and 4 atoms of carbon on right
12 atoms of hydrogen on left and 12 atoms of hydrogen on right
14 atoms of oxygen on left and 14 atoms of oxygen on right
Answer:
We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.
Momentum is a VECTOR quantity having both magnitude and direction. The first ball has momentum P =m*v = 2*4 = 8 at 90degrees. The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees. They sum to zero when you perform vector addition.
Explanation:
The kinetic energy halfway the hill is 
Explanation:
If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:
where
is the initial potential energy, at point A, with
m = 980 kg (mass of the cart)
(acceleration of gravity)
(height at point A)
is the initial kinetic energy, at point A
, with
(velocity at point A)
is the final potential energy, at point B, where
(height at point B)
is the final kinetic energy, at point B, where
is the velocity at point B
Here we are interested in finding 
, so by re-arranging the equation and substituting we find:
Learn more about kinetic energy:
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I would said A is the best option if i’m wrong sorry
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
