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miv72 [106K]
4 years ago
14

according to adam smith and other classical economists, why is the economic theory supporting market economies(or capitalism) mu

ch more realistic than theories supporting sommand economies?
Physics
1 answer:
grin007 [14]4 years ago
3 0
Classical economics is a branch of the economics that states that economy is always moving towards equilibrium, which means that a change in supply will eventually be matched by a change in demand. The father of classical economics  is Adam Smith with his book<span> The Wealth of Nations in 1776.</span>
The economic theory supporting market economies (or capitalism) is much more realistic than theories supporting command economiesIt is based on self interest.
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Zolol [24]

Answer:

(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N

Explanation:

(3) The period of a satellite is given as;

T = 2\pi \sqrt{\frac{r^3}{GM} }

where;

T is the period of the satellite

M is mass of Earth

r is the radius of the orbit

Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

 

(4)

Given;

mass of the ball, m₁ = 1.99 x 10⁴⁰ kg

mass of Neptune, m₂ = 1.03 x 10²⁶ kg

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distance between the Sun and Neptune, r = 4.5 x 10¹² m

The gravitational force between the Sun and Neptune is calculated as;

F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N

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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:
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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

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4 years ago
PLS ANSWER FAST WILL GIVE BRAINLEST!!
egoroff_w [7]

Answer:

from

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m = f/a

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