Which of the following is a part of a wave?

Describe one common critique or criticism of Utilitarianism. To what extent do you agree or disagree with this criticism?

lisabon 2012 [21]

The second most common criticism of utilitarianism is that it is impossible to apply - that happiness (etc) cannot be quantified or measured, that there is no way of calculating a trade-off between intensity and extent, or intensity and probability (etc), or comparing happiness to suffering.

7 0

2 years ago

Using the periodic table, choose the more reactive metal. Al or Mg

ziro4ka [17]

Aluminum, and magnesium are metals. For metals, reactivity decreases as you go from left to right across the periodic table. Atomic number of Al is 13 and of Mg is 12. Hence the least reactive of these two is therefore aluminum.

Magnesium is "HIGHLY FLAMMABLE" carefully take a small piece and hit it with a torch. If its Magnesium it will "Caution, very, quickly burn.
Aluminum will not react to simple flame, it will only melt with enough direct heat.

Magnesium
==========
Atomic Number: 12
Atomic Symbol: Mg
Atomic Weight: 24.305
Electron Configuration: 2-8-2

Aluminum
========
Atomic Number: 13
Atomic Symbol: Al
Atomic Weight: 26.9815
Electron Configuration: 2-8-3

Hope this helps some. Any questions please feel free to ask. Thank you

6 0

3 years ago

Read 2 more answers

The planet Mars has a mass of 6.1 × 1023 kg and radius of 3.4 × 106 m. What is the acceleration of an object in free fall near t

Oksanka [162]

The acceleration due to gravity of Mars is $3.5\ m / s^{2}$

<u>Explanation:</u>

As per universal law of gravity, the gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them. But in the present case, the gravity need to be determined between Mars and the object on Mars. Since the mass of Mars is greater than the mass of any object. Thus,

$\text {Gravitational force of planet}=\frac{G M m}{R^{2}}$

Here, G is the gravitational constant, R is the radius of Mars and M, m is the mass of Mars and the object respectively..

Also, according to Newton’s second law of motion, the acceleration of any object will be equal to the ratio of force exerted on it to the mass of the object.

So in order to determine the acceleration due to gravity of Mars, divide the gravitational force of Mars by mass of object on the surface of Mars.

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{\text {Gravitational force of Mars}}{m \text { of object }}$

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{G M m}{R^{2} \times m}=\frac{G M}{R^{2}}$

$\text { Acceleration due to gravity of mars }=\frac{6.67259 \times 10^{-11} \times 6.1 \times 10^{23}}{3.4 \times 3.4 \times 10^{12}}=\frac{40.703 \times 10^{12}}{11.56 \times 10^{12}}$

$\text { Acceleration }=3.5\ \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago

A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of

Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

1.29 kg/m³ = <u> mass </u>

1800 m³

mass = 1.29 kg/m³ ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0

3 years ago

Read 2 more answers

An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$