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Airida [17]
4 years ago
6

How can the density of an object be determined?”

Physics
2 answers:
Elanso [62]4 years ago
8 0

Answer:

m/p, aka mass/volume.

Explanation:

for example if you had an object that had a mass of 10 and volume of 5, the density of that object would be 10/5, or 2.

Nastasia [14]4 years ago
6 0

Answer:

First you need to know the mass (grams) of the object and its volume, then what you do is  Divide the mass by the volume in order to get an object's Density.

Explanation:

hope that helped good luck :)

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What determines if a material is a conductor or an insulator? Explain your answer and provide at least one example of each.
Dmitry [639]
A conducting material conducts or allows electricity to flow, while an insulator does not allow electricity to flow. For example think of a water pipe, if the pipe has a hole water can flow, on the other hand if it is just a solid rod, no water can flow through. I hope this helps.
4 0
3 years ago
after a large snowstorm you shovel 2000 kilograms of snow off your side walk in 1 hour. you lift the shovel to an average height
solmaris [256]

Lifting a mass to a height, you give it gravitational potential energy of

       (mass) x (gravity) x (height)  joules.

To give it that much energy, that's how much work you do on it.

If 2,000 kg gets lifted to 1.25 meters off the ground, its potential energy is

         (2,000) x (9.8) x (1.25) = 24,500 joules.

If you do it in 1 hour (3,600 seconds), then the average power is

           (24,500 joules) / (3,600 seconds) = 6.8 watts.

None of these figures depends on whether the load gets lifted all at once,
or one shovel at a time, or one flake at a time.

But this certainly is NOT all the work you do.  When you get a shovelful
of snow 1.25 meters off the ground, you don't drop it and walk away, and
it doesn't just float there. You typically toss it, away from where it was laying
and over onto a pile in a place where you don't care if there's a pile of snow
there. In order to toss it, you give it some kinetic energy, so that it'll continue
to sail over to the pile when it leaves the shovel.  All of that kinetic energy
must also come from work that you do ... nobody else is going to take it
from you and toss it onto the pile.


8 0
3 years ago
The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0
3 years ago
Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o
ivolga24 [154]

The correct answer is A.

The coefficient of absorption of material A is 30%. So, the material will absorb 30% energy of the incident wave falling on it. Thus, the reflected wave will carry the rest 70% energy.

The coefficient of absorption of material B is 47%. So, the material will absorb 47% energy of the incident wave falling on it. Thus, the reflected wave will carry the rest 53% energy.

The coefficient of absorption of material C is 62%. So, the material will absorb 62% energy of the incident wave falling on it. Thus, the reflected wave will carry the rest 28% energy.

Hence, material C would be the best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest.

8 0
3 years ago
Read 2 more answers
What is one measurement needed to calculate the speed of an object? Direction Mass Time Velocity
Vlad [161]

The answer is Time.

Formula: Distance/Time

3 0
3 years ago
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