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4 years ago

Which gas has molecules with the greatest average molecular speed at 25°c?

Physics

1 answer:

8_murik_8 [283]4 years ago

7 0

Answer: the options to the questions are

a. 1.0 moles of N2

b.0.5 moles of New

c.0.2 moles of CO2

d.2 moles of He

Answer D

Explanation:

The average molecular speed v of gas is given by =√(8RT,/πM)

From the equation it can be seen that substance with lowest molar mass has the highest velocity has He is the answer

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At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi

Nataliya [291]

Answer:

The distance from the charge is 3.35 m.

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

8 0

3 years ago

The steering wheel of a certain vehicle has a diameter of 34.6 cm, and it turns a shaft that is 8.9 cm in diameter. If a 94.8 N

Vesnalui [34]

Answer: 164Nm

Explanation: the torque exerted on the wheel is given by the expression

Torque=force*radius(of steering wheel )

Given force =94.8N

Diameter of wheel=34.6cm

Radius =34.6/2 =17.3cm

to metre 17.3/10= 1.73m

Torque =94.8*1.73

=164Nm

4 0

3 years ago

Mercury has an average disease to the sun of 0.39 AU. In two or more complete sentences, explain how to calculate the orbital pe

alukav5142 [94]

Answer: 88 Earth days

Explanation:

According to the Kepler Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit:

$T^{2}=a^{3}$ (1)

If we assume the orbit is circular and apply Newton's law of motion and the Universal Law of Gravity we have:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (2)

Where $M$ is the mass of the massive object and $G$ is the universal gravitation constant. If we assume $M$ constant and larger enough to consider $G$ really small, we can write a general form of this law:

$MT^{2}=a^{3}$ (3)

Where $T$ is in units of Earth years, $a$ is in AU (1 Astronomical Unit is the average distane between the Earth and the Sun) and $M$ is the mass of the central object in units of the mass of the Sun.

This means when we are making calculations with planets in our solar system $M=1$ .

Hnece, in the case of Mercury:

$(1)T^{2}=(0.39 AU)^{3}$ (4)

Isolating $T$ :

$T=\sqrt{(0.39 AU)^{3}}$ (5)

$T=0.243 Earth-years \frac{365 days}{1 Earth-year}=88.6 days \approx 88 days$ (6)

This means the period of Mercury is 88 days.

7 0

3 years ago

Two negative charges that are both -3.0 C push each other apart with a force of 19.2 N. How far apart are the two charges ?

fgiga [73]

The Coulomb's force acting between two charges is
$F=k_e \frac{q_1 q_2}{r^2}$
where $k_e = 8.99 \cdot 10^9 N m^{-2}C^{-2}$ is the Coulomb's constant, q1 and q2 the two charges, and r the distance between them.

Using $q_1 = q_2 = -3 C$ , we can find the distance between the two charges when the force is F=19.2 N:
$r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99\cdot 10^9 \frac{(-3.0C)^2}{19.2 N} }=6.5 \cdot 10^4 m = 65 km$

5 0

3 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

leva [86]

Answer:

A. 5.600 m

B. 5.800 s

C. 0.966 m/s

D. 0.315 m

Explanation:

A. The wavelength is the distance between 2 crests, which is 5.600 m

B. Period of the wave is the time for the wave to complete 1 circle (highest point to next highest point). Since it takes 2.9s to travel from highest point to lowest point, it would take another 2.9 to travel from lowest point to the next highest point. So the total time is 2.9 + 2.9 = 5.8 s,

C. The wavespeed is wavelength over unit of time:

5.6 / 5.8 = 0.966 m/s

D. The amplitude would be half the length of highest point to lowest point, which is 0.63 / 2 = 0.315 m

5 0

3 years ago