All categories

2 years ago

A solid cube of aluminum (density of 2.7 g/cm³) has a volume of 0.9 cm³. how many atoms are contained in the cube?

Physics

1 answer:

Reika [66]2 years ago

8 0

Answer:

$0.542*10^{23}\ Aluminum\ Atoms$

Explanation:

$We\ are\ given:\\Density\ of\ aluminum=2.7\ g/cm^3\\Volume\ of\ aluminum-cube=0.9\ cm^3\\Hence,\\As\ we\ know\ that,\\Density=\frac{Mass}{Volume}\\Mass=Density*Volume\\Hence,\ here,\\Mass\ of\ the\ solid\ iron\ cube=2.7*0.9=2.43\ g\\Now,\\We\ also\ know\ that,\\Gram\ Atomic\ mass\ of\ Aluminum = 26.98 \approx 27\ g\\Hence,\\No.\ of\ particles=\frac{Mass}{GAM}*Avagadro's Constant\\Hence,\ here\\No.\ of\ Aluminum\ atoms=\frac{2.43}{27}*6.022*10^{23} \approx 0.542*10^{23}\ Aluminum\ Atoms$

You might be interested in

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

2 years ago

0.75 km expressed in centimeters

disa [49]

75000 lol enjoy..............using up 20 characters

7 0

2 years ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

IgorLugansk [536]

Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Momentum of the t-shirt:

$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .

8 0

1 year ago

Rank the objects below in order from smallest to largest, where 1 is the smallest and 5 is the largest. local group of galaxies

sasho [114]

1. earth
2. solar system
3. milky way galaxy
4. local group of galaxies
5. universe

4 0

3 years ago

A charged particle accelerated to a velocity v enters the chamber of a mass spectrometer. The particle's velocity is perpendicul

gladu [14]

Answer:

Circle

Explanation:

When a charged particle is in motion in a region with magnetic field, the particle experiences a force whose magnitude is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

$\theta$ is the angle between the directions of v and B

In this problem, the velocity of the particle is perpendicular to the magnetic field, so

$\theta=90^{\circ}$

and the formula reduces to

$F=qvB$

Also, the direction of this force is perpendicular to the direction of motion of the particle. This means that as the charge moves in the region of the magnetic field, the force acting on it acts as a centripetal force: therefore, the particle will start moving by unifom circular motion, with constant speed (because the magnetic force does no work on the particle, since it is perpendicular to the direction of motion).

So, the path of the particle will be a circle.

4 0

3 years ago

A solid cube of aluminum (density of 2.7 g/cm³) has a volume of 0.9 cm³. how many atoms are contained in the cube?​

A solid cube of aluminum (density of 2.7 g/cm³) has a volume of 0.9 cm³. how many atoms are contained in the cube?