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GREYUIT [131]
3 years ago
7

What is the benefit of using the metric system and SI Units? Question 1 options: to measure accurately on equipment to convey in

formation accurately so scientists can refer to the same quantity no matter the location or language so that all measurement systems can be used together there is no need to use the same system of measurements around the world
Physics
1 answer:
Kryger [21]3 years ago
4 0

Answer:

The correct option is;

To convey information accurately so scientists can refer to the same quantity no matter the location o language

Explanation:

The International System of Units or SI Units which was established in 1960, consist of the modernized  metric system of units.

The metric system of units came about by the signing of the Metre Convention in 1875 to initiate international cooperation in metrology which initially consisted only of the standards for the measurement of length in metre and the measurement of mass in kilograms. The units for electric currents and other physical units where included in 1875

The SI unit which is founded and also maintained by the General Conference on Weights and Measures (CGPM) is officially recognized by most countries in the world.

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A driver in a car,originally moving 12.2 m/s, applies the brakes until the car comes to a stop. The car moves a distance of 36.5
mojhsa [17]

First we have to calculate the acceleration of the car,

a =\frac{u-v}{t}

Here, u is initial velocity of the car and its value is given 12.2 m/s and v is final velocity of the car and it comes to stop, so its value zero.

a=\frac{0-12.2 m/s}{t} =\frac{-12.2 \ m/s}{t}.

As during the braking the acceleration is constant, from the kinematic equation,

s=ut + \frac{1}{2} a t^2

Here, s is the distance traveled by the car during braking and its value is given 36. 5 m.

Substituting all the values in kinematic equation, we get

36.5 m =(12.2 m/s) t + \frac{1}{2} (- \frac{12.2 m/s}{t}) t^2 \\\\ 36.5 \ m = (12.2 \ m/s) t -(6.1 \ m/s) t \\\\\ t = \frac{36.5 m}{6.1 m/s} = 5.98 s

Therefore, car will stop after 5.98 s.


8 0
3 years ago
Heeeeelp urgent<br><br> I need an compare and contrast between energy and work.<br><br> 20 pts.....
ddd [48]

energy is the ability to do work while work is the dot product of force and displacement

6 0
3 years ago
Read 2 more answers
How would you describe the atmosphere to a friend.
Misha Larkins [42]

Answer:

its a shield that can protect us from super hot rays and all though ita not that strong atleast we wont get burned to death lol :)

6 0
3 years ago
In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040.
lina2011 [118]

Answer:

42.11 years old

Explanation:

Given that:

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040

To find her age we use:

\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\

Δtm is  time interval for the observer stationary relative to the sequence of

events = 2040 - 2000 = 40 years

Δts is is the time interval for an observer moving with a speed v relative to the  sequence of event

v = velocity = 2.5 x 10^8 m/s

c = speed of light = 3 x 10^8 m/s

\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old

7 0
3 years ago
Sketch the electric field around these two objects if they have the same sign of charge. Make a separate drawing showing equipot
diamong [38]

Answer:

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

Explanation:

The electric field around two objects with charge of the same sign, for simplicity suppose that the objects have positive point spherical charges,

          E = k q / r2

bold letters indicate vectors, therefore the total electric field is

           E_total = E1 + E2

the module of this field is

           E_total = E1- E2

therefore we can outline this field

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

An outline of these shows in Attachment A

The equipotential surfaces are defined as being perpendicular to the electric field lines since the electric field and the power difference are related

              E = \frac{dV}{dx} i^ + \frac{dV}{dy} j^ + \frac{dV}{dz} k^ = \Delta V

We can schematize some characteristics of these surfaces

* very close to each load are spherical surfaces

* very far from the load is an elliptical surface, which envelops the loads

* between them there is a point of zero potential point C

See attached part B

5 0
3 years ago
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