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3 years ago

What conventions are used in SI to indicate units

Physics

1 answer:

klemol [59]3 years ago

5 0

Answer:

Conventions used in SI to indicate units are as follows:

Only singular form of units are used. for example: use kg and not kgs.
Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.
Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.
Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.
While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.
Units named after a scientist should be written in small letters. for example: newton, henry.
Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k

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Which of these discoveries is generally attributed to kepler?

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The question is incomplete and no options are given but the answer is;
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3 years ago

A paper clip moves towards a magnet lying on a table. What forces are present in this situation? How do these forces compare?

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2 years ago

A bag of sugar weighs 2 kg on earth. What should it weigh in newtons on the moon, where the free-fall acceleration is 1/6 that o

Elis [28]

Weight = mass x acceleration
Earths acceleration is 9.8 m/s*2
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The mass of the bag at all three locations is 0.91 kg. Mass does not change, the different locations only change its weight.

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3 years ago

Light from a laser of wavelength λ1 shines normally on a pair of narrow slits separated by distance D. This results in a differe

lana66690 [7]

Answer:

Explanation:

The angular position of first intensity maxima is given by the expression

= λ / d where λ is wave length of light and d is slit separation

putting the values given in the problem. ,

.02 = λ₁ / D

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= 2λ₁ /D

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3 years ago

A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b

mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

$V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s$

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

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3 years ago