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sergey [27]
3 years ago
6

Why learning SI base unit are important to understanding the sizes of a different object?

Physics
1 answer:
Flauer [41]3 years ago
7 0
A base unit (also referred to as a fundamental unit) is a unit adopted for measurement of a base quantity. A base quantity is one of a conventionally chosen subset of physical quantities, where no quantity in the subset can be expressed in terms of the others. The International System of Units (SI), commonly known as the metric system, is the international standard for measurement. ... The SI is made up of 7 base units that define the 22 derived units with special names and symbols. The units and their physical quantities are the second for time, the metre for measurement of length, the kilogram for mass, the ampere for electric current, the kelvin for temperature, the mole for amount of substance, and the candela for luminous intensity. The various features and future prospects of the International System of Units (SI) are described. The SI is based on seven selected base units, corresponding to the seven quantities such as length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. Units are of two types: Fundamental units. Derived units. or powering the base units in various combinations, For example:
mechanical work is force applied multiplied by distance moved and has the unit newton metre written as Nm.
speed is distance divided by time and has the unit metre per second written as ms. The three measures are descriptive, diagnostic, and predictive. Descriptive is the most basic form of measurement.
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The area of the effort and load of the Piston of a hydrolic are 0.5 and 5m respectively. If a force of 100 Newton is applied on
Yanka [14]

Answer:

Explanation:

Using Pascal laws, which states that pressure are the input equals the pressure at the output.

Pressure is given as force/area

P1=P2

Then,

F1/A1=F2/A2

Cross multiply

F1A2=F2A1

Given that

Ae=0.5m² area of effort

Al=5m² area of load

Fl=? Force if load

Fe= 100N. Force of effort

Then applying pascal

Fl/Al=Fe/Ae

Fl/5=100/0.5

FL/5=200

Fl=200×5

Fl=1000N

The first safety load is 1000N

6 0
3 years ago
Photographs of many young stars show long jets of material apparently being ejected from their poles.
Rudiy27

Answer:

A) True

Explanation:

Researchers have detected numerous jets of gas ejected from poles of young stars and planetary nebulae.

By examining images of hydrogen molecules excited at infrared wavelengths, scientists have been able to see through the gas and dust in the Milky Way, in order to observe the most distant targets. These goals are normally hidden from view and many of them have never been seen before.

The entire study area covers approximately 1,450 times the size of the full moon, or the equivalent of an image of 95 gigapixels. The survey reveals jets emanating from proto-stars and planetary nebulas, as well as remnants of supernovae, the illuminated edges of vast clouds of gas and dust, and the warm regions that surround massive stars and their associated groups of smaller stars.

7 0
3 years ago
Now switch to “Position vs. Time #2” on the Choose a graph menu. Once again, experiment until you are able to reproduce the grap
seraphim [82]

Answer:

The caterpillar had to move across the horizontal segments. I hope this helps :)

Explanation:

6 0
3 years ago
Read 2 more answers
A ball rolling on grass has more friction than a ball rolling on tile
Dmitry_Shevchenko [17]

Answer:

When you slide an object there is less friction than when you roll it. an object's acceleration (change in speed) depends on the mass of the object and the amount of force applied on it. an object at rest stays at rest unless a force acts on it. ... A ball rolling on grass has more friction than a ball rolling on tile.

6 0
2 years ago
Read 2 more answers
the radius of earth is about 6.38 x10^3 km. A 7.20 x10^3 N spacecraft travels away from earth. What is the weight of the spacecr
statuscvo [17]

Answer:

1796.65\ \text{N}

Explanation:

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

w = Weight of spracecraft at the surface = 7.2\times10^3\ \text{N}

m = Mass of spracecraft

R = Radius of Earth = 6.38\times10^3\ \text{km}

h = Elevation = 6.38\times10^3\ \text{km}

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

M = Mass of Earth = 5.972\times 10^{24}\ \text{kg}

w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{7.2\times 10^3}{9.81}\\\Rightarrow m=733.94\ \text{kg}

From the gravitational law we have

w'=\dfrac{GMm}{(r+h)^2}\\\Rightarrow w'=\dfrac{6.674\times10^{-11}\times 5.972\times 10^{24}\times 733.94}{(6.38\times10^6+6.38\times10^6)^2}\\\Rightarrow w'=1796.65\ \text{N}

The weight of the spacecraft at the given height is 1796.65\ \text{N}

8 0
3 years ago
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