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3 years ago

Write a brief desperation for each of newtons three laws

1 answer:

4 0

1st law: law of inertia, an object will remain stationary unless a force acts upon it.
2nd law: law of acceleration: the momentum experienced by an object is proportional to the size and direction of the force.
3rd law: law of action and reaction: for every action there is an equal and opposite reaction.

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A certain man needed 400N of force to pull a load through a distance of 150cm in 8s . Calculatell i) the work done by man ii) th

Igoryamba

Answer:

a) 600 J

b) 75 W

Explanation:

Force= 400 N, distance = 150 cm = 1.5 m, time= 8s

a) Work is the product of force acting on an object and distance (or displacement). The S.I unit of work is the joules.

Work = force × distance = 400 × 1.5 = 600 J

b) Power is the amount of energy transferred per unit time. It is the ratio of work to time. The S.I unit of power is watt

Power = work/time = 600 / 8 = 75 W

5 0

3 years ago

Pliz, answer my this all question

never [62]

18. a) The materials that are in contact. The two materials and the nature of their surfaces. ...
b) The force pushing the two surfaces together. Pushing the surfaces together causes the more of the asperities to come together and increases the surface area in contact with each other.

19. the quantity of motion of a moving body, measured as a product of its mass and velocity.

20. According to Newton's third law of motion, action force is equal to reaction but acts on two different bodies and in opposite directions. When a horse pushes the ground, the ground reacts and exerts a force on the horse in the forward direction. This force is able to overcome friction force of the cart and it moves.

21. The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.

22. R12. Mass is more fundamental because it is an intrinsic property of an object. Weight varies with location depending upon the acceleration due to gravity eg. for a mass m = 10kg on Earth it`s weight is W = mg = 10 x 10 = 100N.

8 0

3 years ago

Số oxi hóa của các nguyên tố: Fe(OH)2; Fe(OH)3; FexOy; Zn(OH)2 là gì

BARSIC [14]

Answer:

esfsef

Explanation:

esfsefseF

7 0

3 years ago

An acid sample of an unknown concentration is contained in a erlenmeyer flask. Which technique below would be best suited to ana

Ivenika [448]

Answer:

Titration

Explanation:

The best technique which can be used to determine the number of moles of the HCl in the sample is titration.

The given amount of HCl solution must be titrated with known concentration of the base like NaOH.

The volume of NaOH required must be noted also.

According to the reaction,

$NaOH+HCl\rightarrow NaCl+H_2O$

At equivalence point

Moles of $HCl$ = Moles of $NaOH$

Considering:-

Moles of $HCl=Molarity_{NaOH}\times Volume_{NaOH}$

Thus, in this way, moles of HCl can be determined.

4 0

3 years ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 $\frac{cal}{gC}$ .

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

Q is the heat exchanged by a body of mass m.
C is the specific heat substance.
ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

Mass of metal = 50 g
Initial temperature of metal= 45 °C
Final temperature of metal= 11.08 ºC
Specific heat of metal= ?

For water:

Mass of water = 250 g
Initial temperature of water= 10 ºC
Final temperature of water= 11.08 ºC
Specific heat of water = 1.035 $\frac{cal}{gC}$

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 $\frac{cal}{gC}$ × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= $\frac{279.45 cal}{1696 gC}$