how do you calculate the numerical value of physical quantity​

Answers

Answer 1

The value of a physical quantity is normally expressed as an implied product of a numeric value and a unit of measurement.

There are three categories to consider:

There is no explicit unit of measurement included. Examples of this would include index of refraction of a medium and the specific gravity of a substance (which is ratio of the density of the material divided by the density of some reference material, usually water at some specified temperature). In this category, there is an implied measurement unit of 1 . It is usually not written because 1 times any number is that same number, so it is pointless to write the “times 1”. The value of an index of refraction is simply a number, and that number is all you write for the quantity value. That number is the numerical value of the physical quantity. It is only slightly more complicated for specific gravity, because you are dividing one density by another, and both values should be expressed in the same units of measurement, and the division of one by the other cancels out those units, leaving you with 1 as the overall measurement unit.

For plane angles, there is a relationship between the length s of the arc of a circle, the radius r of that circle, and the angle a subtended by the arc at the circle center:

a = s/r

with the angle a being measured in the unit of radians. (To write the formula for some other angular unit requires incorporating a numeric factor, which is basically a conversion factor from radians to degrees,) Thus, if you have a circle of radius 3 m and an arc of 6 m on that circle, the the angle subtended or formed is:

(6 m)/(3 m) = 2, but we said this is the number of radians, so it is 2 rad.

Notice, we are dividing a length by a length (both the arc length and the radius being lengths), so if we use the same measurement unit for both lengths (regardless that unit being meters, feet, parsecs, or anything else), the two units cancel each other out upon division. This means that the unit we are calling radian is like with specific gravity in #1—it has the value 1. Indeed, we see the formula gives us 2 and we know that it is 2 rad, and the only way we can have them be the same, 2 rad = 2, is if the unit radian is actually just a funny, special name for the number 1. Why do we give the number 1 a special name here, unlike in category #1? That is because some inexperienced people find the concept of radian to be strange and inconvenient. They would rather use degrees, or arcminutes, or arcseconds, or semicircles, or some other such unit, and they all have different sizes. For example, a full circle is 2π rad and it is also 360°. Therefore, since both equal one circle of rotation, they must be equal to each other:

360° = 2π rad. Divide both sides by 360 to get:

1° = (2π/360) rad = (π/180) rad. Now, we saw above that rad = 1, so:

1° = (π/180) rad = (π/180) × 1 = π/180.

Thus, like the radian, the degree is also a number—not 1 though, but rather π/180, which cannot be “thrown away” because π/180 times a number does not yield back the original number.

Thus, 30° = 30 × π/180 = π/6 = π/6 × 1 = π/6 rad.

This is the explanation as to when we express an angle in degrees, we must write the ° symbol or spell out degrees, whereas when we express the angle in radians, we may either explicitly write rad or we may leave it off. Unfortunately secondary school geometry textbooks do not seem to understand this point and typically leave off the mandatory ° symbol. That usually gets straightened out when radians are presented—typically later in the second year of algebra or in trigonometry, but it becomes something necessary for students to unlearn the incorrect and learn the correct. Thus, if an angular unit is included, you can convert that angular unit into a real number and multiply by the numeric part of the physical quantity value to the the numeric value of the physical quantity. (And absence of angular unit implies radians, which have numeric value 1, so the numeric value of the quantity is just the numeric value that is present.

Solid angles work similarly, involving area divided by area. The steradian (sr) is the unit that has value 1.


Related Questions

A 0.0250kg golf ball is placed on a spring that is compressed 0.0500m. How high will the ball get when the spring is released if the spring constant is 200.N/m?



20.4m


2.04m


1.02m


40.8m

Answers

Answer:

1.02m

Explanation:

Apply Principle of conversation of energy,

mgh = 1/2 Ke²

Who exerts more pressure? a) A girl of 50 kg, wearing heels with an area of 1 cm2. b) An elephant of 4000 kg with foot area of 250 cm2.

Answers

Answer:

The girl exerts more pressure.

Explanation:

Pressure can be defined as the force exerted normally or perpendicularly per unit area.

i.e P = F/A

Girls

Area of the heel = 1cm² = 10^(-4) m²

Force = mg = 50 × 10 = 500N

Pressure =

[tex] \frac{500}{10 ^{ - 4} } [/tex]

[tex] = 5 \times {10}^{6}[/tex]

Elephant

Area = 250cm² = 2.5 x 10^(-2)b m²

Force = mg = 40000N

Pressure =

[tex] \frac{40000}{2.5 \times {10}^{ - 2} } [/tex]

[tex] = 1.6 \times {10}^{6} [/tex]

Electrons could be swiped off of the
outermost shell of the electron
True
False

Answers

Yeah, it true not false

if you are planning to go for hiking on the mountain,what type of shoes will you wear​

Answers

Answer:

you will have to wear hiking shoes or sports shoes

Worth 10 points! I will give brainliest!! Please help!!! See picture!

Answers

Correct answer is B.

11 An unstretched spring is 12,0 cm long. A load of 5.0N stretches it to 15.0cm. How long will it be under a load
of 15N? (Assume that the spring obeys Hooke's law.)

Answers

Answer: Approximately 22 cm

=========================================================

Explanation:

The unstretched spring is 12.0 cm long. When adding a load of 5.0 N, it stretches to 15.0 cm. This is a displacement of 15.0 - 12.0 = 3.0 cm, which is the amount the spring is stretched.

Convert this displacement to meters (so that it fits with the meters unit buried in Newtons).

3.0 cm = (3.0)/100 = 0.03 m

Apply Hooke's Law to find the spring constant k

F = -kx

5.0 = -k*(0.03)

k = -(5.0)/(0.03)

k = -166.667 approximately

Now we must find the displacement x when F = 15 newtons

F = -kx

-kx = F

x = F/(-k)

x = -F/k

x = -15/(-166.667)

x = 0.089 approximately

x = 0.1

The displacement to one decimal place is about 0.1 meters, which converts to 100*0.1 = 10 cm

So the spring will be stretched to about 12cm+10cm = 22 cm

10. Numerical problems:

b. A stone is thrown vertically upwards with an initial velocity of 20 m/sec.find the maximum height it reaches and the time taken by it to reach the height. (g = 10 m/s2).

c. A car is moving at a rate of 72 km/hr. How far does the car move when it stops after 4 seconds?

d. A bus starts from rest. If the acceleration is 2 m/s2, find
1. the distance travelled.
2. the velocity after 2 seconds.

e. A stone is thrown vertically upward with the velocity of 25 m/s. How longo it take to reach the maximum height? Also calculate the height.


with process please ​

Answers

Answer:

a)20 m,2s

Explanation:

Given

Initial velocity=20 m /sec

g=10 m/s square

v =0m/s

h =?

t =?

now,

v = U +gt

=v=20 +10t

Or,10t=20

Or,t=20

10

therfore ,2 s.

Again,

h=ut -1 gt square

2

Or, h=20×2-1×10(2)square (multiply and cut )2×5 =10

; 2

Or,40-5×4

therefore , 20m

For question e,

u = 25 ms-1
v = 0 m/s
a = -10m/s^2

We have to find time and distance,

For time,

v = u + at

Which can be rephrased as,

v - u/a = t

t = 0 - 25 / -10

t = 2.5 secs

Hence, the stone takes 2.5 secs to reach the maximum height.

To calculate the maximum height attained by the stone, we have to calculate distance/ displacement .

The formula for calculating distance is,

s = ut + 1/2at^2

s = 25 * 2.5 + 1/2 * -10 * 2.5 * 2.5

s = 62.5 + (- 31.25)

s = 62.5 - 31.25

s = 31.25mts

Hence, the maximum distance travelled by the stone is 31.25 meters.

Hope this helps you. Plz let me know if you have any doubts.

An object of mass 2kg is thrown vertically downwards with an initial kinetic energy of 100J. What is the distance fallen by the object at the instant when its kinetic energy has doubled

Answers

The distance fallen by the object at the instant when its kinetic energy has doubled is 5.097m

The kinetic energy of an object is the energy possessed by the body by virtue of its motion.

The gravitational potential energy (energy due to the virtue of its position) of a body is being converted to Kinetic energy once an object falls.

Using the law of conservation of energy according to the equation:

Kinectic Energy = Potential Energy

Since PE = mgh

KE = PE = mgh

Given the following parameters

KE = 100J

m = 2kg

g = 9.81m/s²

Required

Height h

Substitute the required values into the formula

100 = 2(9.81)h

h = 100/19.62

h = 5.097m

Hence the distance fallen by the object at the instant when its kinetic energy has doubled is 5.097m

Learn more here: https://brainly.com/question/8101588

 

People used to believe that if a penny was dropped ff Tom the Empire State Building, how fast would it be moving just before impact on ground? From release of your hand it takes 45 seconds falling at 9.8 m/s^2 due to gravity l.

Answers

Answer:

99

Explanation:

s dispersion dispersion
, refraction

Answers

Answer:

If you want definition, here it is:

Dispersion is defined to be the spreading of white light into its full spectrum of wavelengths. Refraction is responsible for dispersion in rainbows and many other situations. The angle of refraction depends on the index of refraction, as we saw in The Law of Refraction.

The density of gold is 19.3 g cm³. What is the mass of a bar of gold in Kg that measures 6 cm x 4 cmx to 2 cm ?

solve​

Answers

Answer: 0.9264 kg

Explanation: [I'll use "cc" for cubic centimeter, instead of cm^3.

The volume is 6cm*4cm*2cm = 48 cm^3 (cc).

Density of Au is 19.3 g/cc

Mass of gold = (48 cc)*(9.3 g/cc) = 926.4 grams Au

1 kg = 1,000 g

(926.4 grams Au)*(1 kg/1,000 g) = 0.9264 kg, 0.93 kg to 2 sig figs

At gold's current price of $57,500/kg, this bar is worth $53,268. Keep it hidden from your lab partner (and instructor).

Runner A is initially 8.4 km west of a flagpole and is running with a constant velocity of 4.8 km/h due east. Runner B is initially 6.0 km east of the flagpole and is running with a constant velocity of 8.4 km/h due west. What will be the distance of the two runners from the flagpole when their paths cross?
Thanks for your help :)

Answers

Answer:

Distance = 2m

Explanation:

Runner A : 1,75

Runner B : 0.71 ( divided speed over velocity for both & B)

Distance = 1,75 + 0,71 = 2,46 m

Is it right?

Based on the ionic notation shown here, what has happened to this ion. ***List all that apply!!!*** Fe2+ *

Atom has become an anion.

Atom has become a cation.

Atom has lost 2 electrons.

Atom has gained 2 electrons.​

Answers

Answer:

atom has become cation

atom has lost 2 electrons

Explanation:

when atom loses electron they become positively charged and cation are positive ion

a block occupies 0.2587 ft. what is the volume in mm?

Answers

Answer:

70

Explanation:

What is pascal law? Define it​

Answers

Answer:

Pascal's law states that the pressure exerted anywhere in a confined incompressible liquid is transmitted equally in all directions irrespective of the area on which it acts and it always acts at right angles to the surface of containing vessel.

have a great day sister

4) Gabo bought three chicken dinners for $6.95 each and one hamburger meal for $5.75. He and three friends decide to divide the cost of the meals equally. How much money does Gabo receive in total from his three friends?
(A) $6.35 (B) $6.65 (C) $19.65 (D) $19.95 (E) $26.60

Answers

Answer: the answer is d $19.95

Explanation:

chicken dinner: 6.95 x 3 = 20.85

burger: 5.75

20.85 + 5.75 = 26.6

26.6 divided by 4 (bc theres 4 ppl and the bill needs to be split equally)

26.6/4= 6.65

and since 3 friends are paying him back u gotta do 6.65 x 3 = 19.95

Question 1

A car travels 480 km in 12 hr. What is its velocity?

Answers

Explanation:

480km = 480000m

12hr = 43200 sec

480000/43200 = 11.1111111111

round up 2 sig fig

11m/s

answer is 11m/s

how to convert m/s to miles/seconds

Answers

Answer:

Divide by 1609

Explanation:

So to convert from m/s to m/s divide by 1609

A bracelet's mass is 80 g, and its volume is 10 cm3. What is its density?

Answers

Answer:

Density is 8 g/cm³

Explanation:

[tex]density = \frac{mass}{volume} \\ [/tex]

mass » 80 g

volume » 10 cm³

[tex]density = \frac{80 \: g}{10 \: {cm}^{3} } \\ \\ = 8 \: {gcm}^{ - 3} [/tex]

Answer:

mass=80

volume=10

density=mass/volume

which is 80/10

so the density is equals to 10.

can sound waves propagate in vacuum if yes why if no why not??​

Answers

Answer:

please mark me brainliest

Explanation:

No, sound waves cannot travel in a vacuum. This is because sound waves are mechanical waves that need a medium (such as air and water) to travel. Sound in physics is defined as a vibration that propagates as an audible wave of pressure, through a medium such as a gas, liquid or solid.

35/h expressed in meter per second

Answers

Answer:

9.722

Explanation:

35*1000÷1×60*60

35000÷3600

9.722

define null point
define null point ​

Answers

Answer:

In physics a null is a point in a field where the field quantity is zero as the result of two or more opposing quantities completely cancelling each other

how do i solve this?

Answers

Answer:

hmmm i dont know....

Explanation:

i just wanted free point. TANKS YOU SIR!!

For which length of wire are the readings of resistance the most precise

Answers

The resistance of a wire is directly proportional to the length of the wire. That is the longer the length of the wire, the higher the resistance and the shorter the length of the wire, the smaller the resistance.

0.0084x91 ...................

Answers

Answer:

0.0084×91=0.7644

Explanation:

hope this helps

0.0084×91=0.7644

please mark this answer as brainlist

what happens when you try to walk on the smooth well polished floor wearing brand new shoes​

Answers

You will skip since there is no friction between the smooth floor and the smooth soles of the new shoes.

How do particles differ after a physical change

Answers

Answer:

In physical changes, no new materials are formed and the particles do not change apart from gaining or losing energy. ... Particles stay the same unless there is a chemical change whether the matter is solid, liquid or gas. Only their arrangement, energy and movement change.

Explanation:

A motorcycle is moving with the velocity of 72km/h. If the velocity reaches 40m/s after 0.5m calculate the acceleration

Answers

Answer:

72km/hr

= 72× 1000m/1×60×60 sec

= 72000m/ 3600sec

=20m/s

0.5 m

= 0.5×60 sec

= 30 sec

Now,

acceleration = v-u/t

= 40m/s-20m/s divided by 30 sec

= 20m/s divided by 30 sec

= 0.6m/s2

Explanation:

hope this helps you

The quantity of matter inside the body is?

Answers

Answer:

mass is the quantity of matter inside the body

Answer:

The mass of a body is the quality of matter it contains

Plzz help me fast

(There is both physics and chemistry question )

1) Write the molecular formula of the following by crisscross method :
1. Potassium sulphate
2. Sodium chloride
3. Ammonium chloride
4. Silver oxide


2) Prove that F=ma pl

Answers

Potassium Sulphate :-

[tex]\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,3){\bf K}\put(0,0){\bf 1}\put(4,3){\sf SO_4}\put(4,0){\bf -2}\put(0.2,2.9){\vector(4,-3){3.5}}\put(3.8,2.9){\vector(-4,-3){3.5}}\end{picture}[/tex]

[tex]\boxed{\sf K_2SO_4}[/tex]

Sodium chloride:-

[tex]\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,3){\bf Na}\put(0,0){\bf 1}\put(4,3){\sf Cl}\put(4,0){\bf -1}\put(0.2,2.9){\vector(4,-3){3.5}}\put(3.8,2.9){\vector(-4,-3){3.5}}\end{picture}[/tex]

[tex]\boxed{\sf NaCl}[/tex]

Ammonium Chloride

[tex]\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,3){\bf NH_4}\put(0,0){\bf 1}\put(4,3){\sf Cl}\put(4,0){\bf -1}\put(0.2,2.9){\vector(4,-3){3.5}}\put(3.8,2.9){\vector(-4,-3){3.5}}\end{picture}[/tex]

[tex]\boxed{\sf NH_4Cl}[/tex]

Silver Oxide

[tex]\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,3){\bf Ag}\put(0,0){\bf 1}\put(4,3){\sf O}\put(4,0){\bf -2}\put(0.2,2.9){\vector(4,-3){3.5}}\put(3.8,2.9){\vector(-4,-3){3.5}}\end{picture}[/tex]

[tex]\boxed{\sf Ag_2O}[/tex]

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