Which particles account for most of the mass of an atom?

Vaselesa [24]

Answer:

The first law, also called the law of inertia, was pioneered by Galileo. This was quite a conceptual leap because it was not possible in Galileo's time to observe a moving object without at least some frictional forces dragging against the motion. In fact, for over a thousand years before Galileo, educated individuals believed Aristotle's formulation that, wherever there is motion, there is an external force producing that motion.

The second law, $ f(t)=m\,a(t)$ , actually implies the first law, since when $ f(t)=0$ (no applied force), the acceleration $ a(t)$ is zero, implying a constant velocity $ v(t)$ . (The velocity is simply the integral with respect to time of $ a(t)={\dot v}(t)$ .)

Newton's third law implies conservation of momentum [138]. It can also be seen as following from the second law: When one object ``pushes'' a second object at some (massless) point of contact using an applied force, there must be an equal and opposite force from the second object that cancels the applied force. Otherwise, there would be a nonzero net force on a massless point which, by the second law, would accelerate the point of contact by an infinite amount.

Explanation:

7 0

3 years ago

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Who can do my worksheet for me

timurjin [86]

Answer:

what is it on? like name one of the questions

Explanation:

5 0

3 years ago

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

4 years ago

Use what you know about reflection and absorption of light to complete the sentence.

Blababa [14]

A red ladybug appears red in white light, red in red light, and black in blue light. Those would be the proper selections you'd need.

3 0

3 years ago

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When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the

Aleksandr [31]

Answer:

Tangential

Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.

7 0

3 years ago