A measure of the pull of gravity on an object.

Easy defimation of newtons second law of motion

shtirl [24]

Hi pupil here's your answer ::

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Newton's Second Law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
ie., F=ma

Where F is the force applied, m is the mass of the body, and a, the acceleration produced.

Or in simplest language it is the force applied to a particular object of particular mass multiplied by the acceleration caused by force .

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hope that it helps. . . . . .

8 0

3 years ago

Which process requires more energy: completely vaporizing 1 kg of saturated liquid water at 1 atm pressure or completely vaporiz

Lilit [14]

The correct answer would be the first option. The process that would need more energy would be vaporizing 1 kg of saturated liquid water at a pressure of 1 atmosphere. This can be seen from the latent heat of vaporization of each system. For the saturated water at 1 atm, the latent heat is equal to 40.7 kJ per mole while, at 8 atm, the latent heat is equal to 36.4 kJ per mole. The latent heat of vaporization is the amount of heat needed in order to vaporize a specific amount of substance without any change in the temperature. As we can observe, more energy is needed by the liquid water at 1 atm.

7 0

3 years ago

You need to repair a gate on the farm. The gate weighs 100 kg and pivots as indicated. A small diagonal bar supports the gate an

tekilochka [14]

Answer:

The force is $F = 3920 \ N$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The weight of the gate is $G = 100\ kg$

The vertical component of F is $F_y = F\ sin \theta$

From the diagram , taking moment about the pivot we have

$W_g * 2 - F_y * 1 = 0$

Where $W_g$ is the weight of the gate evaluated as

$W_g = m_g * g = 100 * 9.8 = 980 \ N$

=> $980 * 2 - Fsin(30) * 1 = 0$

=> $F = \frac{1960}{sin(30)}$

=> $F = 3920 \ N$

7 0

3 years ago

g We saw in class that in a pendulum the string does no work. We also saw that the normal force does no work on an object slidin

Ymorist [56]

Answer:

From question (a) and (b) the pendulum motion is perpendicular to the force so the normal force will do no work and the tension in the string of the pendulum will not work

$i.e Normal \ Force(N) = mg cos \theta$

And $\theta = 90$ so

$N = 0$

c

An example will be a where a stone is attached to the end of a string and is made to move in a circular motion while keeping the other end of the string in a fixed position

d

A dog walking along a surface which has friction, here the frictional force would acting in the direction of the motion and this would do positive work

Explanation:

5 0

3 years ago

Read 2 more answers

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0

3 years ago