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2 years ago

The newton is defined as the:

Physics

2 answers:

natta225 [31]2 years ago

8 0

Answer: C

Explanation:

irakobra [83]2 years ago

4 0

The Newton is defined as the:

C. Force that can give a 1-kilogram mass an acceleration of 1m/sec squared.

<h3>Hope it helps..</h3>

ray4918 here

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3 years ago

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3 years ago

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Determine the components of reaction at the fixed support

denis-greek [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are

$A_{(x)} = 400 \ N$ , $A_{(y)} = -500 \ N$ , $A_{(z)} = 600 \ N$ , $M_x = 1225 \ N\cdot m$ , $M_y = 750 \ N\cdot m$ , $M_z = 0 \ N\cdot m$

Explanation:

Looking at the diagram uploaded we see that there are two forces acting along the x-axis on the fixed support

These force are 400 N and $A_{(x)}$ [ i.e the reactive force of 400 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(x)} - 400 = 0$

=> $A_{(x)} = 400 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the y-axis on the fixed support

These force are 500 N and $A_{(y)}$ [ i.e the force acting along the same direction with 500 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(y)} + 500 = 0$

=> $A_{(y)} = -500 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the z-axis on the fixed support

These force are 600 N and $A_{(z)}$ [ i.e the reactive force of 600 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(z)} - 600 = 0$

=> $A_{(z)} = 600 \ N$

Generally taking moment about A along the x-axis we have that

$\sum M_x = M_x - 500 (0.75 + 0.5) + 600 ( 1 ) = 0$

=> $M_x = 1225 \ N\cdot m$

Generally taking moment about A along the y-axis we have that

$\sum M_y = M_y - 400 (0.75 ) + 600 ( 0.75 ) = 0$

=> $M_y = 750 \ N\cdot m$

Generally taking moment about A along the z-axis we have that

$\sum M_z = M_z = 0$

=> $M_z = 0 \ N\cdot m$

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3 years ago

Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

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3 years ago