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3 years ago

Stade avogadro's hypothesis what are its applications, prove that hydrogen hydrogen and oxygen gases

Physics

1 answer:

natita [175]3 years ago

3 0

Answer:

Applications of Avogadro's hypothesis:

In explaining Gay Lussac's law of gaseous volumes.

In determining the atomicity of gasses.

In determining the molecular formula of a gas.

In establishing the relationship between relative molecular mass and vapor density.

Explanation:

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Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter

kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

$T_e=2\pi \sqrt{\frac{L}{g_e}}$

where

$g_e = 9.8 m/s^2$ is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

$T_m=2\pi \sqrt{\frac{L}{g_m}}$

where

$g_m = 1.6 m/s^2$ is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

$\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}$

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are 15.0 cm apart, the repulsive force between them h

Tems11 [23]

Answer:

a) q_1=q_2= 7.42*10^-7 C

b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C

Explanation:

Given:

F_e = 0.220 N

separation between spheres r = 0.15 m

Electrostatic constant k = 8.99*10^9

Find: charge on each sphere

part a

q_1 = q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_1^2 = F_e*r^2/k

q_1=q_2= sqrt (F_e*r^2/k)

Plug in the values and evaluate:

q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)

q_1=q_2= 7.42*10^-7 C

part b

q_1 = 4*q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_2^2 = F_e*r^2/4*k

q_2= sqrt (F_e*r^2/4*k)

Plug in the values and evaluate:

q_2= sqrt (0.22*0.15^2/4*8.99*10^9)

q_2= 3.7102*10^-7 C

q_1 = 14.8*10^-7 C

4 0

3 years ago

Read 2 more answers

Which nucleus completes the following equation?

melomori [17]

The nucleus that completes the equation given above is 218/85 At, which is option D.

<h3>What is a nuclear fission?</h3>

Nuclear fission is a nuclear reaction in which a large nucleus splits into smaller ones with the simultaneous release of energy.

According to this question, the following nuclear fission reaction is given: 222/87Fr = 4/2He+ ?

To find the missing nucleus, we subtract 4 from 222 to get 218, therefore, the nucleus that completes the equation given above is 218/85 At.

Learn more about nuclear fission at: brainly.com/question/913303

#SPJ1

4 0

2 years ago

A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil h

tresset_1 [31]

Answer:

The current pass through the coil is 6.25 A

Explanation:

Given that,

Diameter = 25 cm

Magnetic field = 1.0 mT

Number of turns = 100

We need to calculate the current

Using the formula of magnetic field

$B =\dfrac{\mu_{0}NI}{2\pi r}$

$I=\dfrac{B\times2\pi r}{\mu N}$

Where, N = number of turns

r = radius

I = current

Put the value into the formula

$I=\dfrac{1.0\times10^{-3}\times2\pi\times12.5\times10^{-2}}{4\pi\times10^{-7}100}$

$I=6.25\ A$

Hence, The current passes through the coil is 6.25 A

6 0

3 years ago

A typical ten-pound car wheel has a moment of inertia of about 0.35kg⋅m2. The wheel rotates about the axle at a constant angular

QveST [7]

Answer:

The rotational kinetic energy of the rotating wheel is 529.09 J

Explanation:

Given;

moment of inertia I = 0.35kg⋅m²

number of revolutions = 35.0

time of revolution, t = 4.00 s

Angular speed (in revolution per second), ω = 35/4 = 8.75 rev/s

Angular speed (in radian per second), ω = 8.75 rev/s x 2π = 54.985 rad/s

Rotational kinetic energy, K = ¹/₂Iω²

Rotational kinetic energy, K = ¹/₂ x 0.35 x (54.985)²

Rotational kinetic energy, K = 529.09 J

Therefore, the rotational kinetic energy of the rotating wheel is 529.09 J

7 0

3 years ago

Stade avogadro's hypothesis what are its applications, prove that hydrogen hydrogen and oxygen gases​

Stade avogadro's hypothesis what are its applications, prove that hydrogen hydrogen and oxygen gases