if the pressure on a gas increases, the volume of the gas also increases is that statement true or false?

A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an

Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

6 0

3 years ago

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________ occurs when the horizontal winds move air against mountain ranges, forcing air to rise as it passes over the mountains.

Semenov [28]

The answer would be:
Precipitation sometimes occurs when the horizontal winds move air against mountain ranges, forcing air to rise as it passes over the mountains.
This happens when the air is forced to move from low elevation to high elevation due to rising terrain. This causes the air to cool adiabatically. This increases the relative humidity and causes clouds to form, under certain conditions it can also create precipitation.

6 0

3 years ago

Please Help!

e-lub [12.9K]

Answer:

Q9. Man who received the most altercations for a theory which later on became a revolutionary theory influenced in many areas of modern science and technology.

Q10. Fire truck is coming towards you

Explanation:

Q9. Christian Doppler was born on 29th of November 1803 in Saltzburg. After studies in Linz and Vienna, he graduated in Mathematics. For many years, Doppler struggled to find work in academia, and for a time he worked as a bookkeeper at a factory. His academic career took him from Austria to Prague, where he became assistant at the University and later worked as professor in Prague. Back to Vienna, he was appointed as professor at the Polytechnic School and in 1850 as first director of the new Institute of Physics. While working at Vienna, his health broke down and moved Venice where he sought his eternal rest on March 17th, 1953.

During his lifetime, the man was quite controversial: a personality praised by some, but detested by others; and even as a scientist, he had a difficult time. He did publish papers on magnetism, electricity, optics and astronomy but, the discovery that allowed him to remain in history of science was the one he presented at Royal Bohemian Society of Science entitled "On the colored light of the double stars and certain other stars of the heavens" in 1842. He hypothesized that the pitch of the sound would change if the source was moving.

Doppler's ideas were initially received with a certain amount of skepticism so, in order to support his claims, he devised an experiment in 1845 with the help of colleague. He used two sets of trumpeters, one set stationary at a train station and one set moving on an open train car. Both sets of musicians had perfect pitch and held the same note. As the train passed the station, it was obvious that the frequency of the two notes didn't match, even though the musicians were playing same note. This proved his hypothesis.

Demonstrating that the Doppler effect also held true for frequency of ligh proved more difficult and was never successfully achieved before Doppler's demise. The first experiment that revealed a Doppler shift in starlight was carried out at the beginning of twentieth century. Since then Doppler effect was proved invaluable for astronomical observations.

For the most of the academic world, he is known as physicist; but one can equally find him on the list of mathematicians and astronomers too. This is proof for the exceptional broad spectrum of application of his main discovery.

Q10. When there is increase in frequency of the sound from source, then the source is moving towards you. Hence the fire truck is coming towards you

3 0

3 years ago

A chemical equation lists NaHCO3(s) as a reactant. What does (s) indicate?

Ostrovityanka [42]

Answer:

The (s) indicates that the state of matter for NaHCO3 is solid.

Explanation:

When a chemical reaction is written, the state of matter for each components of the reactants and products are mentioned in brackets along with their names or formulas.

For example, NaHCO3 has (s) mentioned in the brackets. The s shows that the state of matter for NaHCO3. (l) represents liquid format. (g) represents that the state of matter is gas.

5 0

3 years ago

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A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago