What the opposite of suspension physical science physical science

Calculate the number of atoms contained in a cylinder (1 m radiusand1 m deep)of (a) magnesium (b) lead.

Jobisdone [24]

Answer:

The question is incomplete,below is the complete question

"Calculate the number of atoms contained in a cylinder (1μm radius and 1μm deep)of (a) magnesium (b) lead."

Answer:

a. 1.35*10^{11} atoms

b. 1.03*10^{11} atoms

Explanation:

First, we determine the volume of the magnesium in the cylinder container

using the volume of a cylinder

$V=\pi r^{2}h\\ r=10^{-6}m\\ h=10^{-6}m\\V=\pi *10^{-6*2}*10^{-6}\\V=\pi *10^{-18}\\V=3.14*10^{-18}m^{3}\\$

a. Next we determine the mass of the magnesium ,

using the density=mass/volume

since density of a magnesium

$the density of magnesium =1.738*10^{3}kg/m^{3} \\mass=density * volume \\mass=1.738*10^{3}*3.14*10^{-18}\\mass=5.46*10^{-15}kg\\ \\mass=5.46*10^{-12}g\\$

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

$mole=5.46*10^{-12}/ 24.305\\mole=0.225*10^{-12}mol\\$

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

$number of atoms = 0.225*10^{-12}*6.02*10^{23}\\number of atoms =1.35*10^{11} atoms$

b. for he lead, we determine the mass of the lead ,

using the density=mass/volume

since density of a magnesium

$the density of lead =11.34*10^{3}kg/m^{3} \\mass=density * volume \\mass=11.34*10^{3}*3.14*10^{-18}\\mass=35.60*10^{-15}kg\\ \\mass=35.60*10^{-12}g\\$

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

$mole=35.60*10^{-12}/ 207.2\\mole=0.1718*10^{-12}mol\\$

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

$number of atoms = 0.1718*10^{-12}*6.02*10^{23}\\number of atoms =1.03*10^{11} atoms$

3 0

2 years ago

the eccentricity of the Moon's orbit is low, medium, or high with respect to most of the planets' orbits around the sun?

12345 [234]

person sorry answering for points

4 0

3 years ago

A helium weather balloon is filled to a volume of 219 m3 on the ground, where the pressure is 754 torr and the temperature is 29

Novosadov [1.4K]

Answer:

$V_2 = 606.9 m^3$

Explanation:

By ideal gas equation law we know that

PV = nRT

now we know that when balloon rises to certain level then the number of moles will remains same

so we can say

$n_1 = n_2$

$\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}$

now plug in all data to find the final volume of the balloon

$\frac{754\times 219}{R(298)} = \frac{210 \times V_2}{R(230)}$

$V_2 = \frac{230\times 754 \times 219}{210 \times 298}$

$V_2 = 606.9 m^3$

8 0

2 years ago

At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertica

Diano4ka-milaya [45]

Answer:

(a). Energy is 64,680 J

(b) velocity is 51.43m/s

(c) velocity in mph is 115.0mph

Explanation:

(a).

The potential energy $P$ of the payload of mass $m$ is at a vertical distance $h$ is

$P =mgh$ .

Therefore, for the payload of mass $m = 50kg$ at a vertical distance of $h = 132 m$ , the potential energy is

$P = (50kg)(9.8m/s^2)(132m)$

$\boxed{P = 64,680J}$

(b).

When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,

$mgh= \dfrac{1}{2}mv^2$

$v= \sqrt{2gh}$

$v = \sqrt{2*9.8*135}$

$\boxed{v = 51.43m/s}$

(c).

The velocity in mph is

$\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}$

$\boxed{v= 115.0mph}$

5 0

3 years ago

Once a baseball has been hit into the air, what forces are acting upon it? How can you tell that any forces are acting upon the

Tom [10]

Well, its in the air, so the air is "upon" the ball. and when it comes down...you catch it, and throw it, and get someone out, and win the game, and just keep doing that, and boooommm you're and pro baseball player. Life is good

8 0

3 years ago