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3 years ago

What is the primary atomic characteristic that Mendeleev used to organize the periodic table?

Physics

2 answers:

Fudgin [204]3 years ago

7 0

Atomic number is the answer

Svet_ta [14]3 years ago

6 0

Atomic mass is the primary atomic characteristic that Mendeleev used to organize the periodic table.

The atomic mass (ma) is the mass of an atomic particle, sub-atomic particle, or molecule. It is commonly expressed in unified atomic mass units (u) where by international agreement, 1 unified atomic mass unit is defined as 1/12 of the mass of a single carbon-12 atom (at rest).

The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Help me please I need to turn it in before midnight...

ale4655 [162]

Answer:

The statement which explains how the total time spent in the air is affected as the projectile's angle of launch increases from 25 degrees to 50 degrees is;

C. Increasing the angle from 25° to 50° will increase the total time spent in the air

Explanation:

The equation that can be used to find the total time, T, spent in the air of a projectile is given as follows;

$T = \dfrac{2 \cdot u \cdot sin(\theta) }{g}$

Where;

T = The time of flight of the projectile = The time spent in the air

u = The initial velocity of the projectile

θ = The angle of launch of the projectile

g = The acceleration due to gravity ≈ 9.81 m/s²

Given that sin(50°) > sin(25°), when the angle of launch, θ, is increased from 25 degrees to 50 degrees, we have;

Let T₁ represent the time spent in the air when the angle of launch is 25°, and let T₂ represent the time spent in the air when the angle of launch is 50°, we have;

$T_1 = \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}$

$T_2 = \dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g}$

sin(50°) > sin(25°), therefore, we have;

$\dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g} > \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}$

Therefore;

T₂ > T₁

Therefore, increasing the angle at which the projectile is launched from 25° to 50° will increase the total time spent in the air.

7 0

3 years ago

A student is at rest on a stool that may freely spin about its central axis of rotation. As the stool spins, the student holds o

steposvetlana [31]

Answer:

the final speed of the stool is 3.6 rad/s

Explanation:

As we know that there is no external torque on the system

So we can use concept of angular momentum conservation

So we will have

$I_1\omega_1 = I_2\omega_2$

now we will have

$6\times 1.2 = 2\times \omega$

$\omega = \frac{6 \times 1.2}{2}$

$\omega = 3.6 rad/s$

So the final speed of the stool is 3.6 rad/s

8 0

3 years ago

Which value is equivalent to 7.2 kilograms?

emmainna [20.7K]

7200 grams is what equivalen to 7.s kilograms

3 0

3 years ago

A star may form a black hole once it cannot ?

IceJOKER [234]

A star may form a black hole once is true. After the marked supernova explosion, the star shrinks because it sheds away the gaseous layers of the star. And especially if the star is large, it will form into a black hole.

4 0

3 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

$F_{No}= 7771.125 \ N$

$F_p = 2.2*10^{6} N$

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

4 years ago