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tatuchka [14]
3 years ago
7

WHAT IS 6 1/12 - 4 2/3=

Mathematics
1 answer:
Ede4ka [16]3 years ago
3 0
6 1/12 - 4 2/3
6 1/12 - 4 8/12
5 13/12 - 4 8/12
1 5/12
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The (Figure 1) shows two thin beams joined at right angles. The vertical beam is 19.0 kg and 1.00 m long and the horizontal beam
il63 [147K]

A) The center of gravity of the two joined beams is;

<u><em>(x, y) = (25/44), (19/88)</em></u>

B) The gravitational torque on the joined beams about an axis through the corner is;

<u><em>τ = 245 N.m</em></u>

We are given;

Mass of vertical beam; m₂ = 19 kg

Mass of horizontal beam, m₁ = 25 kg

Length of horizontal beam; L₁ = 2 m

Length of vertical beam; L₂ = 1 m

A) Formula for center of gravity of the two joined beams is;

(x, y) = [(m₁x₁ + m₂x₂)/(m₁ + m₂)],  [(m₁y₁ + m₂y₂)/(m₁ + m₂)]

Where;

(x₁, y₁) is the <em>center of gravity</em> on the <em>horizontal beam</em>

(x₂, y₂) is the <em>center of gravity</em> on the <em>vertical beam</em>

(x₁, y₁) = (L₁/2, 0)

Plugging in the relevant values gives;

(x₁, y₁) = (2/2, 0)

(x₁, y₁) = (1, 0)

(x₂, y₂) = (0, L₂/2)

Plugging in the relevant values gives;

(x₂, y₂) = (0, 1/2)

(x₂, y₂) = (0, 0.5)

Thus, center of gravity of the 2 joined beams is;

(x, y) = [((25 × 1) + (19 × 0))/(25 + 19)],  [((25 × 0)  + (19 × 0.5)))/(25 + 19)]

(x, y) = (25/44), (19/88)

B) Formula for the gravitational torque on the joined beams about an axis through the corner is given as;

τ = (m₁g)x₁ + (m₂g)x₂

Plugging in the relevant values;

τ = (25 × 9.8)1 + (19 × 9.8)0

τ = 245 N.m

Read more at; brainly.com/question/14825257

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2 years ago
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Lol seems like elementary but 17.50 minus 9.50 equals 8 so the basic change is 8. X'D I cant believe this is college X'D
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Why is the end behavior of a quadratic function different from a linear function?
inna [77]

A linear function can’t repeat but a quadratic function can

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John has 1 chicken jake took one how much does john have left
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Answer:

0 because 1-1=0 and that is the answer

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