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3 years ago

Write the importance of international bureau of weight and measures in the country

Physics

1 answer:

MArishka [77]3 years ago

6 0

Answer:

International Bureau of Weights and Measures (BIPM) is an international organization founded to bring about the unification of measurement systems, to establish and preserve fundamental international standards and prototypes, to verify national standards, and to determine fundamental physical constants. The bureau was established by a convention signed in Paris on May 20, 1875, effective January 1876. In 1921 a modified convention was signed. The convention provides for a General Conference that meets every four years to consider required improvements or modifications in standards. An International Committee of Weights and Measures, composed of 18 scientists elected by the conference, meets annually to monitor worldwide uniformity in units of measure.

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A dog runs 30 feet to the north then 5 feet to the south what is the displacement of the dog

Nuetrik [128]

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The freight cars a and b have a mass of 20 mg and 15 mg, respectively. if the cars collide and couple together, what is the velo

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3 years ago

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4 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$