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3 years ago

12

When plugging in metric facts, always remember that 1 big unit = # small units. Fill in these facts: 1. s= μs 2.

____1____ g = 0.001________ kg 3._______ km = ________m 4.________mm = _______ m 5.________mL = ______ L 6._________g = ______ dg 7.________cm = _______ m 8.________ms = ________s

1 answer:

balandron [24]3 years ago

4 0

Answer:

1. 1 s = 1 x 10⁶ μs

2. 1 g = 0.001 kg

3. 1 km = 1000 m

4. 1 mm = 1 x 10⁻³ m

5. 1 mL = 1 x 10⁻³ L

6. 1 g = 100 dg

7. 1 cm = 1 x 10⁻² m

8. 1 ms = 1 x 10⁻³ s

Explanation:

1.

1 x 10⁻⁶ s = 1 μs

(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs

<u>1 s = 1 x 10⁶ μs</u>

2.

1000 g = 1 kg

1 g = 1/1000 kg

<u>1 g = 0.001 kg</u>

3.

<u>1 km = 1000 m</u>

<u></u>

4.

<u>1 mm = 1 x 10⁻³ m</u>

<u></u>

5.

<u>1 mL = 1 x 10⁻³ L</u>

<u></u>

6.

1 x 10⁻² g = 1 dg

(1 x 10⁻² x 10²) g = 1 x 10² dg

<u>1 g = 100 dg</u>

<u></u>

7.

<u>1 cm = 1 x 10⁻² m</u>

<u></u>

8.

<u>1 ms = 1 x 10⁻³ s</u>

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This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is $A_{x}$ = 0

the y-component of the force at A is $A_{y}$ = 400 N

the couple acting at A is; $M_{A}$ = 146 N-m

b)

the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑ $M_{L}$ = 0

Vector force acting at A, $F_{A}$ = $A_{x}i$ + $A_{y}j$

Now, From the image, we have;

a)

∑f = 0

$F_{A}$ - 600j + 200j = 0i + 0j

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$A_{x}i$ + ( $A_{y}$ - 400)j = 0i + 0j

now by equating i- coefficients'

$A_{x}$ = 0

so, the x-component of the force at A is $A_{x}$ = 0

also by equating j-coefficient

$A_{y}$ - 400 = 0

$A_{y}$ = 400 N

hence, the y-component of the force at A is $A_{y}$ = 400 N

we also have;

∑ $M_{L}$ = 0

$M_{A}$ - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

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$M_{A}$ = 146 N-m

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The sum of the moments about right end of the beam is;

∑ $M_{R}$ = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)( $A_{y}$ ) + $M_{A}$

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

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M = -(30 N-m) + (112 N-m) - ( 228 N-m))

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