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3 years ago

How is budgeting different for most states than for the federal government?

Physics

1 answer:

meriva3 years ago

5 0

There are three general kinds of state balanced budget requirements:

The governor's proposed budget must be balanced (43 states and Puerto Rico).The budget the legislature passes must be balanced (39 states and Puerto Rico).The budget must be balanced at the end of a fiscal year or biennium, so that no deficit can be carried forward (37 states and Puerto Rico).
Nature of state balanced-budget requirements All the states except Vermont have a legal requirement of a balanced budget. Some are constitutional, some are statutory, and some have been derived by judicial decision from constitutional provisions about state indebtedness that do not, on their face, call for a balanced budget. The General Accounting Office has commented that "some balanced budget requirements are based on interpretations of state constitutions and statutes rather than on an explicit statement that the state must have a balanced budget."

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Which statement best describes how light waves travel in a uniform medium?

Zanzabum

The statement which best describes how light waves travel in an uniform medium is in straight lines. The correct answer will be A.

5 0

3 years ago

Read 2 more answers

For a wire has a circular cross section with a radius of 1.23mm.

Mila [183]

Answer:

$5.731\times 10^{-5}\ m/s$

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = Conduction electron density in copper = $8.49\times 10^{28}\ electrons/m^3$

$v_d$ = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

$I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s$

The drift speed of the electrons is $5.731\times 10^{-5}\ m/s$

$v_d=\dfrac{I}{neA}$

From the equation we can see the following

$v_d\propto \dfrac{1}{n}$

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

5 0

3 years ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

3 years ago

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

umka21 [38]

Answer:

Explanation:

s = s₀ + v₀t + ½at²

s = 0 + 0(15) + ½(6)(15²)

s = 675 m

Not sure what the free fall acceleration is needed for, but if the object is dropped from a high enough point, it will travel in 15 seconds

s = ½10(15²) = 2250 m if air resistance is ignored

7 0

3 years ago