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tino4ka555 [31]
3 years ago
9

Performing Metric Unit Conversions

Physics
2 answers:
Umnica [9.8K]3 years ago
8 0

Answer:

5000 millimeters

Explanation:

1 meter=100 centimeters

50 centimeter=0.5 meters❌

1 meter = 10 decimeters

500 decimeters = 50 meters❌

1 meter =1000 milimeters

5000 milimeters = 5 meters✔

Anastasy [175]3 years ago
3 0

Answer:

5000 millimeters

Explanation:

1 meter=100 centimeters

50 centimeter=0.5 meters❌

1 meter = 10 decimeters

500 decimeters = 50 meters❌

1 meter =1000 milimeters

5000 milimeters = 5 meters✔

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Explanation:

The X-component of the velocity = Vcosx. Where, V = magnitude of the velocity. The x component of velocity will depend on the diagram. It the angle is measured from the x-axis which is considered the horizontal then Vx = Vcos(theta). The magnitudes of the components of velocity v → are v x = v cos θ and v y = v sin θ , v x = v cos θ and v y = v sin θ , where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in Figure 4.12. Derivation of the Trajectory Formula.

y = refers to the vertical position of the object in meters. x = refers to the horizontal position of the object in meters. Horizontal velocity component: Vx = V * cos(α)

Vertical velocity component: Vy = V * sin(α)

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Four masses are located in the xy plane. A 4.00 kg mass is located at the origin, and a second 4.00 kg mass is located at the po
Sergio [31]

Answer:

A)    I_total = 16 m, B)   I_total = 8 m,  C)   I_total = 8 m, D)  I_total = 8 m

Explanation:

The moment of inertia is a scalar quantity, therefore the total moment of inertia

          I_total = I₁ + I₂ + I₃ + I₄

the moment of inertia of a point mass with respect to an axis of rotation

         I = m r²

Let's apply this to our case

A) Rotation axis at the origin

     I₁ = m 0 = 0

for the second masses, we find the distance using the Pythagorean theorem

     r = \sqrt{2^2 + 2^2}

     r = 2 √2

     I₂ = m (2 √2) ²

     I₂ = 8 m  

     I₃ = m 2² = 4 m

     I₄ = m 2² = 4 m

we substitute

     I_total = 0 + 8m + 4m + 4m

     I_total = 16 m

B) axis of rotation in the center of the square

let's find the distance to any mass

     r = \sqrt{1^2+ 1^2}

     r = √2

     I₁ = m 2

     I₂ = m 2

     i₃ = m 3

     I₄ = m 4

we substitute

    I_total = 4 (2m)

    I_total = 8 m

C) axis of rotation is the x axis

       I₁ = 0

       I₂ = m 2² = 4 m

       I₃ = m 2² = 4 m

       I₄ = 0

       I_total = 8 m

D) axis of rotation is the y-axis

       I₁ = 0

       I₂ = 4m

       I₃ = 0

       I₄ = 4 m

       I_total = 8 m

5 0
3 years ago
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