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3 years ago

Define the difference between the rigid body problem and a single particle problem.

1 answer:

3 0

. In single particle problem whole mass is concentrated at a single point so it has a single displacement, single velocity and single acceleration. while, in rigid body mass is distributed

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The U. S. Navy is helping these Congolese soldiers conduct an inspection of vehicles, looking mostly for car bombs. They can ins

dangina [55]

Interesting question. I'd take a guess that it could be either mirrors and light travelling in straight lines and being reflected, or optical fibres to achieve a perhaps similar result.

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3 years ago

Read 2 more answers

Which equation best summarizes Newton’s 2nd law:

zepelin [54]

Option B makes best sense, correct me if i’m wrong

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3 years ago

Lawson's criterion states that the product of ion density and confinement time must exceed a certain number before a break-even

LekaFEV [45]

According to Lawson's criterion, the outcome is determined by the product of ion density and confinement time because the temperature must be maintained for a sufficient confinement time and with a sufficient ion thickness to obtain a net gain of power from a fusion reaction.

<h3>What are Lawson's criterion?</h3>

The overall conditions that must be met in order to produce more energy than is required for plasma heating are usually expressed in terms of the product of ion density and confinement time, a condition known as Lawson's criterion.

In nuclear fusion devices, confinement time is defined as the amount of time the plasma is kept at a temperature above the critical ignition temperature.

Even at temperatures high enough to overcome the coulomb barrier to nuclear fusion, a critical density of ions must be maintained in order to achieve a net yield of energy from the reaction.

Because the density required for a net energy yield is correlated with the confinement time for hot plasma, the minimum condition for a productive fusion reaction is typically stated in terms of the product of ion density and confinement time, which is known as Lawson's criterion.

To learn more about Lawson's criterion, refer:

brainly.com/question/28303495

#SPJ4

7 0

1 year ago

A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washe

STALIN [3.7K]

To solve this exercise it is necessary to take into account the concepts related to thermal expansion.

The thermal expansion is given by the function,

$\Delta L = L_0 \alpha \Delta T$

Where,

$\Delta L=$ Change in Length

$\Delta T =$ Change in Temperature

$\alpha =$ Coeficiente de dilatación lineal

$L_0 =$ Initial Length

By quickly deducing the formula, we can realize that the greater the change in temperature, the greater the change in the length of the radius.

The change in length is proportional to the change in temperature. Considering that the other two terms are constant we have that the correct one would be: <em>The hole in the center of the washer will expand.</em>

4 0

3 years ago

A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra

son4ous [18]

Answer:

The answer is $v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}$ .

Explanation:

From the law of gravity,

$F = G \frac{Mm}{r^2}$

considering F as a conservative force, $F = - \nabla U$ ,

the general expression for gravitational potential energy is

$U = -G \frac{Mm}{r}$ ,

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

$g = -G \frac{M}{r^2}$ ,

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

$U = g M_e R$ ,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

$U_p = g M_e(R- r_s)$ ,

then, the kinetik energy when the rock hits the surface is

$U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s)$ ,

$v = \sqrt{2g(R-r_s)}$

where g is the gravitational acceleration generated by the Sun at R,

$g = G \frac{M_s}{R^2}$ .

8 0

2 years ago