A > Is the correct answer :)
<span>The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16
Traveling against the current:
(18 - c)*t = 8
Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16
Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16
(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0
Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c
So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3
(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3
And it's verified.</span>
Divide 24 by 12.
24/t = 12
24/12 = t
24/12 = 2
t = 2
Answer:
True I hope you like it
Give me a brainliest answer
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
