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3 years ago

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A 0.5 m3 container is filled with a fluid whose specific volume is 0.001 m3/kg. At standard gravitational acceleration, the cont

ents of this container weigh:

1 answer:

xenn [34]3 years ago

5 0

Answer: the contents of this container weighs 4905 kg.m/s²

Explanation:

Given that;

volume of a container V = 0.5 m³

we know that standard gravitational acceleration g = 9.81 m/s²

specific volume of liquid filled in the container v = 0.001 m³/kg

now we express the equation for weight of the container.

W = mg

W = (pV)g

W = Vg / ν

so we substitute

W = (0.5 m³)(9.81 m/s ) / 0.001 m³/kg

W = 4.905 / 0.001

W = 4905 kg.m/s²

Therefore, the contents of this container weighs 4905 kg.m/s²

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If a person is standing erect and flexes the trunk on the hip, the center of mass will move __________ and the line of gravity m

marta [7]

Answer:

anterior

anterior

Explanation:

In the given question is asked that

If a person is standing erect and flexes the trunk on the hip, the center of mass will move ___________ and the line of gravity moves___________ within the base of support.

The current answer to the blanks will be

anterior

anterior

hope this helps any further query can be asked in comment section.

8 0

3 years ago

A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y

Genrish500 [490]

Answer:

Positive z direction.

Explanation:

The magnetic force acting on the electron is given by the formula as :

$F=q(v\times B)$

q is the charge on proton

v is the speed of proton

B is the magnetic field

It is mentioned that the proton is moving with a velocity in the positive x-direction. The uniform magnetic field B in the positive y-direction such taht,

q = +e

v = vi

B = Bj

$F=e(v\ i\times B\ j)$

Since, $i\times j=k$

$F=(evB)\ k$

So, the magnetic force acting on the proton in positive z axis. Hence, the correct option is (d) "positive z direction".

8 0

3 years ago

What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 1

-BARSIC- [3]

Answer:

404K

Explanation:

Data given, Kinetic Energy.K.E=8.37*10^-21J

Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.

This relationship can be expressed as

$K.E=\frac{3}{2}KT$

where K is a constant of value 1.38*10^-23

Hence if we substitute the values, we arrive at

$T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K$

converting to degree we have $131^{0}C$

4 0

3 years ago

What's the voltage drop running through the parallel portion of the circuit?

Leni [432]

R 3/4 = (R3 * R4) / (R3 + R 4) = ( 9 * 18 ) /(9 + 18 ) = 162 / 27 = 6 Ohms
R e = R 1 + R 2 + R 3/4 + R 5 = 3 + 6 + 6 + 15 = 30 Ohms
I = U / Re = 90 V / 30 Ohms = 3 A
Finally for the voltage U 3/4 ( the parallel portion of the circuit ):
U 3/4 = 6 Ohms * 3 A = 18 V
Answer: 18 V

4 0

3 years ago

Read 2 more answers

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>

This is what Einstein proposed:

Light behaves like a stream of particles called photons with an energy $E$ :

$E=h.f$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi$ is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago