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zhenek [66]
3 years ago
15

Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of

Physics
1 answer:
goldfiish [28.3K]3 years ago
5 0

<u>Answer</u>

The combined displacement is 2km north


<u>Explanation</u>


Since displacement is a vector quantity, we take into account the direction.


Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.


We now add to obtain,

4+-2+5+-5

This will simplify to

=4-2+5-5=2

Therefore the combined displacement is 2km north

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Whats the answer???????????
Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

R_{1}=4\Omega

R_{2}=6\Omega

R_{3}=8\Omega

Now, when the resistors are connected in parallel, the total resistance R is calculated as follows:

\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}

Isolating R:

R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}

Rewriting with th known values:

R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}

Finally:

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Hence, the correct option is less than 4 ohms.

4 0
3 years ago
In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue
agasfer [191]

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

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