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Musya8 [376]
3 years ago
9

Which question cannot be answered using scientific methods?

Physics
2 answers:
Ray Of Light [21]3 years ago
8 0

Extra Comment: Judging by the way this question has been worded, I may be answering this question incorrectly, but I’ll try my best to meet what you were asking for!

Answer: Some people say that everything can be answered if you just add a little bit of scientific methods to answering a question, but that is wrong. Not everything can be answered with using scientific methods. For example, questions that can not be answered using any form of a scientific method would be any questions regarding anything personal, such as your name, birthday, or the things that you enjoy. For instance, I like anime, but you can’t figure out that I like anime with science.

serg [7]3 years ago
3 0

Answer:

Questions that cannot be answered through scientific investigation are those that relate to personal preference, moral values, the supernatural, or unmeasurable phenomena.

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What form of government concentrates power in a central government
spin [16.1K]
The answer is
Unitary system : All government power is held by a central government
8 0
3 years ago
Read 2 more answers
Gravitational notes of physics ​
Pachacha [2.7K]

Answer:

Every object in the universe attracts other object by a force of attraction, called gravitation, which is directly proportional to the product of masses of the objects and inversely proportional to the square of distance between them. This is called Law of Gravitation or Universal Law of Gravitation.

Let masses (M) and (m) of two objects are distance (d) apart. Let F be the attractional force between two masses.

Importance of The Universal Law of Gravitation

It binds us to the earth.

It is responsible for the motion of the moon around the earth.

It is responsible for the motion of planets around the Sun.

Gravitational force of moon causes tides in seas on earth.

Free Fall

When an object falls from any height under the influence of gravitational force only, it is known as free fall.

Acceleration Due to Gravity

When an object falls towards the earth there is a change in its acceleration due to the gravitational force of the earth. So this acceleration is called acceleration due to gravity.

The acceleration due to gravity is denoted by g.

The unit of g is same as the unit of acceleration, i.e., ms−2

Mathematical Expression for g

From the second law of motion, force is the product of mass and acceleration.

F = ma

For free fall, acceleration is replaced by acceleration due to gravity.

Therefore, force becomes:

F = mg ….(i)

But from Universal Law of Gravitation,

Factors Affecting the Value of g

As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator.

As we go at large heights, value of g decreases.

To Calculate the Value of g

Value of universal gravitational constant, G = 6.7 × 10–11 N m2/ kg2,

Mass of the earth, M = 6 × 1024 kg, and

Radius of the earth, R = 6.4 × 106 m

Putting all these values in equation (iii), we get:

Thus, the value of acceleration due to gravity of the earth, g = 9.8 m/s2.

Difference between Gravitation Constant (G) and Gravitational Acceleration (g)

S. No.

Gravitation Constant (G)

Gravitational acceleration (g)

1.

Its value is 6.67×10-11Nm2/kg2.

Its value is 9.8 m/s2.

2.

It is a scalar quantity.

It is a vactor quantity.

3.

Its value remains constant always and everywhere.

Its value varies at various places.

4.

Its unit is Nm2/kg2.

Its unit is m/s2.

Motion of Objects Under the Influence of Gravitational Force of the Earth

Let an object is falling towards earth with initial velocity u. Let its velocity, under the effect of gravitational acceleration g, changes to v after covering the height h in time t.

Then the three equations of motion can be represented as:

Velocity (v) after t seconds, v = u + ght

Height covered in t seconds, h = ut + ½gt2

Relation between v and u excluding t, v2 = u2 + 2gh

The value of g is taken as positive in case of the object is moving towards earth and taken as negative in case of the object is thrown in opposite direction of the earth.

Mass & weight

Mass (m)

The mass of a body is the quantity of matter contained in it.

Mass is a scalar quantity which has only magnitude but no direction.

Mass of a body always remains constant and does not change from place to place.

SI unit of mass is kilogram (kg).

Mass of a body can never be zero.

Weight (W)

The force with which an object is attracted towards the centre of the earth, is called the weight of the object.

Now, Force = m × a

But in case of earth, a = g

∴ F = m × g

But the force of attraction of earth on an object is called its weight (W).

∴ W = mg

As weight always acts vertically downwards, therefore, weight has both magnitude and direction and thus it is a vector quantity.

The weight of a body changes from place to place, depending on mass of object.

The SI unit of weight is Newton.

Weight of the object becomes zero if g is zero.

Weight of an Object on the Surface of Moon

Mass of an object is same on earth as well as on moon. But weight is different.

Weight of an object is given as,

Hence, weight of the object on the moon = (1/6) × its weight on the earth.

Try the following questions:

Q1. State the universal law of gravitation.

Q2. When we move from the poles to the equator, the value of g decreases. Why?

Q3. If two stones of 150 gm and 500 gm are dropped from a height, which stone will reach the surface of the earth first and why ?

Q4. Differentiate between weight and mass.

Q5. Why is the weight of an object on the moon 1/6th its weight on the earth??

7 0
3 years ago
Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

5 0
3 years ago
A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.
Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = 0.5mv^{2}

           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
  • when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = 0.5x25x5.6^{2} - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

5 0
3 years ago
A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
aleksandr82 [10.1K]

Answer:

A(t) = -340e^{-t/70} + 350

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

A(0) = 10 g

Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

y = ce^{bt} + \frac{a}{b}

So A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350

with A(0) = 10

c + 350 = 10

c = 10 - 350 = -340

A(t) = -340e^{-t/70} + 350

4 0
3 years ago
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