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denpristay [2]
3 years ago
8

Which of the following is an example of a physical change but not a chemical change?

Physics
2 answers:
vazorg [7]3 years ago
7 0

A water pipe freezes and cracks on a cold night. <em> (D)</em>

There are three physical changes going on in this scenario:

1).  The air gets cold and dark at night.

2).  The water in the pipe freezes and expands.

3).  The pipe cracks.

There are no chemical changes in the description.

Radda [10]3 years ago
5 0

Answer:

<u>A water pipe freezes and cracks on a cold night</u>

Explanation:

A water pipe freezing and cracking is a physical change, not a chemical change. A physical change is a change in what the item looks like, feels like, or how the item has a different form. The water pipe freezing is an example of a physical change because the liquid water has turned to frozen water. The frozen water has expanded the pipe, causing it to crack. There are no chemical changes taking place in this scenario.

Mordancy.

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A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat
forsale [732]

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, a=9.8\ m/s^2

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

F=mg-\dfrac{mv^2}{r}

F=m(g-\dfrac{v^2}{r})

F=30\times (9.8-\dfrac{(7.3)^2}{15})

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

5 0
3 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

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3 years ago
Which of Earth’s spheres is composed of a mixture of gases?
vlabodo [156]

Answer:

Earth's atmosphere is a layer of gases surrounding the planet Earth and retained by the Earth's gravity. It contains roughly 78% nitrogen and 21% oxygen 0.97% argon and carbon dioxide 0.04% trace amounts of other gases, and water vapor. This mixture of gases is commonly known as air.  Explanation:

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3 years ago
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Characteristics of deposition?
goldenfox [79]

Answer:

<em>I hope this helps with your question. :)</em>

Explanation:

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3 years ago
A 2.5 kg , 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turnt
emmainna [20.7K]

The concepts required to solve this problem are those related to the conservation of the angular momentum and the moment of inertia of the disk. We will begin by calculating the moment of inertia of the disc, then the moment of inertia of the disc after the two two blocks hits and sticks to the edges of the turn table. In the end we will apply the conservation theorem.

The radius is given as,

R = \frac{20cm}{2} = 10cm = 0.1m

When a block falls from above and sticks to the turn table, the moment inertia of the turntable increases.

Since two blocks are stick to the turn table, the total final moment of inertia of the turntable is the sum moment of inertias of individual turntable, and two blocks.

I_1 = \frac{1}{2} MR^2

I_1 = \frac{1}{2} (2.5)(0.1)^2

I_1 = 0.0125kg \cdot m^2

The moment of inertia of each block is

I_0 = mR^2

Total moment of inertia of two block is

I_0' = 2mR^2

The final moment of inertia of the turn table is

I_2 = I_1 +I'0

I_2 = I_1 +2mR^2

I_2 = 0.01kg\cdot m^2 + 2(500*10^{-3}kg)(0.1m)^2

I_2 = 0.0225kg\cdot m^2

From the conservation of the angular momentum, the initial angular momentum of the system is equal to final angular momentum of the system,

Rearrange the equation we have that

I_1\omega_2 = I_2\omega_2

\omega_2 = \frac{I_1\omega_2}{I_2}

\omega_2 = \frac{0.01*150rpm}{0.0225}

\omega_2 = 66.67rpm

The magnitude of the turntable's angular velocity is 66.67rpm

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3 years ago
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