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1 year ago

I NEED HELP URGENTLY! FOR 20 POINTS!

Physics

1 answer:

puteri [66]1 year ago

5 0

The law of definite proportion states that any sample of the compound will always have the same percentage of each element by weight or mass.

There are four elements that made up matter: air, water earth, and fire is not the main idea of dalton's atomic theory.

<h3>What is the law of definite proportion?</h3>

The Law of Definite Proportion defines that the elements that composed a chemical compound are generally arranged in a definite mass ratio regardless of preparation.

The composition of compounds will always be the same by mass using the law of definite proportions. Chemical compounds are made of elements that are present at fixed ratios in terms of their mass.

In the nitrogen dioxide molecule (NO₂), Nitrogen and oxygen atoms are always present in a 1:2 ratio.

According to Dalton's atomic theory, all matter consists of atoms. The atoms of the same element have the same mass and the atoms of different have different masses and compounds are made of atoms of more than one element.

Therefore, for question (1) option (C) and for (2) option (C) is correct.

Learn more about the law of definite proportion, here:

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Meteoric material dates the formation of the solar system at about_____ billion years.

Alja [10]

According to meteoric material, the solar system was formed around <u>4.6 billion years</u> ago.

<h3>What substance makes up a meteorite?</h3>

With just trace levels of sulphide and carbide minerals, they are primarily composed of iron-nickel metal.
Many asteroids melted during the radioactive element decay in the early solar system, and the iron they carried, being dense, sank to the centre to create a metallic core.

<h3>What is the term for meteorite metal?</h3>

Meteoric iron, also known as meteoritic iron, is a native metal and a protoplanetary-disk remnant from the early universe that is found in meteorites.
It is mostly composed of the metals iron and nickel, primarily in the crystalline phases kamacite and taenite.

learn more about meteorite here

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6 0

1 year ago

A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff

Nikolay [14]

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ( $E$ ) of the project is equal to the sum of gravitational potential energy ( $U_{g}$ ) and translational kinetic energy ( $K$ ), all measured in joules:

$E = U_{g} + K$ (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}$ (Eq. 1b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y$ - Initial height of the projectile above ground, measured in meters.

$v$ - Initial speed of the projectile, measured in meters per second.

If we know that $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y = 132\,m$ and $v = 126\,\frac{m}{s}$ , the initial mechanical energy of the earth-projectile system is:

$E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}$

$E = 498556.296\,J$

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

$W_{loss} = E_{o}-E_{1}$ (Eq. 2)

Where:

$E_{o}$ - Initial total mechanical energy, measured in joules.

$E_{1}$ - FInal total mechanical energy, measured in joules.

$W_{loss}$ - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$W_{loss} = E_{o}-K_{1}-U_{g,1}$

$W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}$ (Eq. 2b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ - Maximum height of the projectile above ground, measured in meters.

$v_{1}$ - Current speed of the projectile, measured in meters per second.

If we know that $E_{o} = 498556.296\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 297\,m$ and $v_{1} = 89.3\,\frac{m}{s}$ , the work losses due to air friction are:

$W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)$

$W_{loss} = 125960.4\,J$

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

$E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}$ (Eq. 3)

$K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}$

Where:

$E_{1}$ - Total mechanical energy of the projectile at maximum height, measured in joules.

$U_{g,2}$ - Potential gravitational energy of the projectile, measured in joules.

$K_{2}$ - Kinetic energy of the projectile, measured in joules.

$W_{loss}$ - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}$ (Eq. 3b)

$m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}$

$v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}$

$v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }$

If we know that $E_{1} = 372595.896\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{2} =0\,m$ and $W_{loss} = 125960.4\,J$ , the final speed of the projectile is:

$v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }$

$v_{2} \approx 82.475\,\frac{m}{s}$

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

7 0

3 years ago

When cold water is poured into a glass, the glass becomes colder. What type of energy transfer

romanna [79]

Answer:
Thermal energy by conduction

Explanation:
When cold water is poured into a glass, we have two objects in contact each with a different temperature. Therefore, the system is.not at equilibrium.
For the system to reach equilibrium, heat always transfers from the object with high temperature (glass in our case) to the object with less temperature (cold water in our case). This transfer keeps occurring until both objects have the same temperature.

Now, since the two objects are in direct contact, therefore, heat would transfer by conduction

Hope this helps :)

7 0

3 years ago

For general projectile motion, the horizontal component

Aleksandr [31]

Answer:

Zero.

Explanation:

In the general projectile motion there is only one component of acceleration which is in the vertical direction.There is no any acceleration in the horizontal direction.As we know that horizontal component of the velocity is remains constant in the projectile motion.

Therefore we cans say that in the projectile motion ,the acceleration in the horizontal direction is zero.

Therefore the answer will be zero.

4 0

4 years ago

1. Explain how this picture is an example of kinetic energy.

kobusy [5.1K]

Answer:

1. He is putting force on the ball to throw it. 2. Him throwing the ball. 3. If he puts more force into throwing the ball it would give the ball more energy>>>

Explanation:

8 0

4 years ago