Elements in Group 18, at the far right of the periodic table, are called what?

Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea

dimaraw [331]

Answer:

stone A is diamond.

Explanation:

given,

Volume of the two stone = 0.15 cm³

Mass of stone A = 0.52 g

Mass of stone B = 0.42 g

Density of the diamond = 3.5 g/cm³

So, to find which stone is gold we have to calculate the density of both the stone.

We know,

$density$ $\density = \dfrac{mass}{volume}$

density of stone A

$\rho_A = \dfrac{0.52}{0.15}$

$\rho_A = 3.467\ g/cm^3$

density of stone B.

$\rho_B = \dfrac{0.42}{0.15}$

$\rho_B = 2.8\ g/cm^3$

Hence, the density of the stone A is the equal to Diamond then stone A is diamond.

6 0

2 years ago

If the same pressure is exerted over a greater area will more of less force result? Group of answer choices :

Zigmanuir [339]

Answer:

More force

Explanation:

Pressure and force are related by the equation:

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area

We can re-arrange the equation as

$F=pA$

In this problem, the pressure is kept the same (p' = p) while the area is increased. As we can see from the previous equation, the force applied is directly proportional to the area: therefore, a greater area means also a greater force.

8 0

2 years ago

Bullets spin when shot from a rifle or handgun. What causes this spinning?

tamaranim1 [39]

Answer:

The spark from the primer ignites the gunpowder. Gas converted from the burning powder rapidly expands in the cartridge. The expanding gas forces the bullet out of the cartridge and down the barrel with great speed. The rifling in the barrel causes the bullet to spin as it travels out of the barrel.

Explanation:

7 0

2 years ago

A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of

nexus9112 [7]

Answer:

$F_1=366.67Hz$

Explanation:

From the question we are told that:

Frequency of 3rd harmonics $F_3=1100$

Frequency of 5th harmonics $F_3=1833$

Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

$\lambda_3=\frac{4}{3}l$

Therefore

$F_3=\frac{3v}{4l}$

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

$\lambda_1=\frac{4}{1}l$

Therefore

$F_1=\frac{v}{4l}$

Generally the equation for the frequency of the first harmonic is mathematically given by

$F_1=\frac{F_3}{3}$

$F_1=\frac{1100}{3}$

$F_1=366.67Hz$

7 0

2 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: