PHYSICS HIGH SCHOOL

List 3 examples behaviors of solid and the 3nwrgy associated with them

Answers

Answer 1

Answer:

Answer:

retains a fixed volume and shape

rigid - particles locked into place

not easily compressible

little free space between particles

does not flow easily

rigid - particles cannot move/slide past one another

Explanation:

Answer 2

Answer: Sure! Here are three examples of behaviors of solids and their associated energies:

1. Behavior: Solids have a definite shape and volume.
Energy: Solids possess potential energy due to the arrangement of their particles.

2. Behavior: Solids are rigid and maintain their shape when subjected to external forces.
Energy: Solids exhibit elastic potential energy when deformed and can store mechanical energy.

3. Behavior: Solids have a fixed melting point at which they transition to a liquid state.
Energy: Solids require thermal energy (heat) to overcome intermolecular forces and transition to a liquid state.

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Help!!!!1. A material that can be hit without shattering is?
viscous, flammable, hard, malleable
2. At room temp., a substance with a melting point of 40% C is a?
solid, liquid, gas, mixture
3. what action involves a chemical change?
making ice cubes,adding sugar to tea, cutting wrapping paper, baking a cake
4.a substance that has a little tendency to change into other substance is said to have low?
reactivity, viscosity, density, conductivity
5. formation of a precipitate is usually evidence of?
the separation of a mixture, a chemical change, formation of a mixture, physical change

Answers

1. For the first one if it doesn't shatter then it is hard.
2. We sould consider that room temperature is usually from 18°C (64°F) to 23°C (73°F) and since the melting point of the substance is 40˚C the substance is liquid, because is melting point not an evaporation point it just melts from solid to liquid.
3. Well, in this one we have to consider that a chemical change it change the thing completely. So in this case is baking a cake because the products make the cake stop being them self and created something else.
4. It tends to have low reactivity. This is because viscosity means something like honey very sticky , density is how hard it is and lastly the conductivity I would believe is to pass electricity or current.
5. It is a physical change because the evaporated water transforms again into water and then fall as rain, snow or ice.
Hope it helped <3

How does a stars life begin?

Answers

A stars life begins as a nebula (which would mostly be hydrogen) and starts to concentrate and compact with dust which then by gravity turns into a star.

A speedy rabbit is hopping to the right with a velocity of 4.0 \,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction when it sees a carrot in the distance. The rabbit speeds up to its maximum velocity of 13 \,\dfrac{\text m}{\text s}13 s m 13, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction with a constant acceleration of 2.0 \,\dfrac{\text m}{\text s^2}2.0 s 2 m 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction rightward.

Answers

Answer: 38.25 m

Explanation:

In this situation we need to find the distance $d$ between the rabbit and the carrot, and we can use the following equation, since the rabbit's acceleration is constant:

$V^(2)=V_(o)^(2) + 2ad$ (1)

Where:

$V=13 m/s$ is the rabbit's maximum velocity (final velocity)

$V_(o)=4 m/s$ is the rabbit's initial velocity

$a=2 m/s^(2)$ is the rabbit's acceleration

$d$ is the distance between the rabbit and the carrot

Isolating $d$ :

$d=(V^(2)-V_(o)^(2))/(2a)$ (2)

$d=((13 m/s)^(2)-(4 m/s)^(2))/(2(2 m/s^(2)))$ (3)

Finally:

$d=38.25 m$

Answer:

4.5s

Explanation:

Cause that's what it says on my test hints

A bicyclist bikes the 56 mi to a city averaging a certain speed. The return trip is made at a speed that is 6 mph slower. Total time for the round trip is 11 hr. Find the bicyclist's average speed on each part of the trip.

Answers

Answer:

Speed of the bicyclist when going to city = 14 miles per hour.

Speed while return trip = 8 miles per hour.

Explanation:

Let the speed of the bicyclist when going to city = x miles per hour.

Speed while return trip = x - 6 miles per hour.

Total time taken = 11 hrs = Time for the trip to city + time taken for return trip.

Also, Time = Distance / Time.

So,

56 / x + 56 / ( x -6) = 11

11x² -178x + 336 = 0

Solving for x we get:

Acceptable x = 14 miles per hour.

Speed while return trip = x - 6 miles per hour = 8 miles per hour.

Final answer:

To find the average speed on each part of the trip, use the formula Average speed = Total distance/Total time and set up an equation to solve for the unknown speeds.

Explanation:

To find the average speed on each part of the trip, we can use the formula Average speed = Total distance/Total time. Let's assume the average speed on the first part of the trip (56 miles) is x mph. Since the return trip is made at a speed that is 6 mph slower, the average speed on the second part of the trip is (x - 6) mph. We know that the total time for the round trip is 11 hours. So, we can set up the equation:

56/x + 56/(x - 6) = 11

Now, we can solve this equation to find the value of x, which represents the average speed on the first part of the trip. Once we have x, we can find the average speed on the second part of the trip by subtracting 6 from x.

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The Ptolemaic theory of the universe was disproved bya. Descartes.
b. Kepler.
c. Copernicus.
d. Newton.

Answers

The Ptolemaic theory of the universe wasdisproved by Nicolaus Copernicus (1473–1543), letter C and followed by JohannesKepler (1571–1630). Geocentrictheory suggested that the Sun and moon orbited Earth, and the rest of theplanets orbited the Sun is Heliocentric theory. Heliocentric theory of NicolausCopernicus during the Scientific Revolution on 17th and 18thcentury while Geocentric theory was proposed by Ptolemy (AD 150) and Aristotle.Today, we believe more on the Heliocentric theory and it has been proven aswell.

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficient of static and kinetic friction between the box and the level bed of the truck are 0.300 and 0.200, respectively. What is the greatest acceleration that the truck can have before the toolbox slides out?

Answers

Final answer:

The greatest acceleration that the truck can have before the toolbox slides out can be calculated by understanding the balance between the inertia force experienced by the toolbox due to acceleration (F = ma) and the maximum static friction force (fs(max) = μsN) opposing this motion. The truck can accelerate up to the point at which these two forces are equal.

Explanation:

The question relates to a concept in Physics known as Friction. In this scenario, the toolbox on the truck experiences static friction which keeps it from sliding. The maximum force of static friction can be calculated using the equation fs(max) = μsN, where μs is the coefficient of static friction and N is the normal force. In this case, μs is given as 0.300 and the normal force N equals the weight of the toolbox. The truck can accelerate up to the point where the frictional force equals the force caused by acceleration, which is calculated using the equation F = ma, where m is mass and a is acceleration.

When the truck accelerates, an inertia force acts on the toolbox in the opposite direction. This inertia force, F = ma, should not exceed the maximum static friction force, fs(max), otherwise, the toolbox will slide. Hence, with given values of static friction coefficient and mass of the toolbox, the greatest acceleration of the truck to prevent slipping can be calculated by equating the frictional force and inertia force.

Learn more about Friction here:

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Final answer:

The greatest acceleration that the truck can have before the toolbox slides out is 5.00 m/s².

Explanation:

The greatest acceleration that the truck can have before the toolbox slides out can be found by comparing the force of static friction to the force pushing the toolbox forward. In this case, the force of static friction must be equal to or greater than the force pushing the toolbox, which is the product of the mass of the toolbox and its acceleration. Given the coefficient of static friction of 0.300, the maximum force of static friction can be calculated. Using the equation fs <= μsN, where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force, we can substitute the values and solve for the maximum force of static friction which is 196 N. The maximum force of static friction is equal to the product of the mass of the toolbox and its acceleration, which gives us the equation fs = max = (50.0 kg)(5.00 m/s²) = 250 N. Therefore, the greatest acceleration that the truck can have before the toolbox slides out is 5.00 m/s².