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Marina CMI [18]
3 years ago
5

A student is given a sample of matter and asked to determine whether it is an element.

Physics
2 answers:
Pepsi [2]3 years ago
5 0
I think the correct answer is C
scoundrel [369]3 years ago
4 0
It's a mixture cause the 3 different pure substances can only be separated if its a mixture
You might be interested in
A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric
netineya [11]

Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N

F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N

W_{F_x} = F_xd = 227.5\times 4.8 = 1092J

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

W_g = -mg\sin(30^\circ)d

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

W_g = -823J

The energy dissipated by the frictional force can be found as follows:

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N

W_F = Fd = 227.5\times 4.8 = 1092J

W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

8 0
4 years ago
What do we need to generate electricity?
Alekssandra [29.7K]

Answer:

Electricity is most often generated at a power plant by electromagnetic generators.

4 0
3 years ago
Professor blossom expects her students to be on time for class holds, them for the entire class period and has firm deadlines se
defon

Answer:

A dictatorship or tyranny.

Explanation:

8 0
3 years ago
A metal block has a density of 5000 kg per cubic meter and a mass of 15,000 kg. What is its volume?
Naily [24]

Taking into account the definition of density, the volume of the metal block is 3 m³.

<h3>What is density</h3>

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

<h3>Volume of the metal block</h3>

In this case, you know that:

  • Density= 5000 \frac{kg}{m^{3} }
  • Mass= 15000 kg
  • Volume= ?

Replacing in the definition of density:

5000 \frac{kg}{m^{3} } =\frac{15000 kg}{volume}

Solving:

volume×5000 \frac{kg}{m^{3} }= 15000 kg

volume= \frac{15000 kg}{5000 \frac{kg}{m^{3} }}

<u><em>volume= 3 m³</em></u>

In summary, the volume of the metal block is 3 m³.

Learn more about density:

brainly.com/question/952755

brainly.com/question/1462554

#SPJ12

4 0
2 years ago
G of potassium reacts with 16 g of oxygen to produce 94 g of potassium oxide
Vanyuwa [196]

Answer:

78g

Explanation:

Given parameters:

Mass of oxygen gas = 16g

Mass of potassium oxide = 94g

Unknown:

Mass of reacting potassium = ?

Solution:

To solve this problem, we need to obtain a balanced reaction equation. Then determine the number of moles of the reactant and use it to find that of the other one.

Balanced equation:

                      4K +   O₂   →    2K₂O

  Number of moles of reacting oxygen;

       Number of moles = \frac{mass}{molar mass}

molar mass of  O₂ = 2 x 16 = 32g/mole

     Number of moles = \frac{16}{32}  = 0.5mole

From the reaction equation;

          4 mole of K reacted with 1 mole of O₂;

           x mole of K will react with 0.5 mole of O₂

Therefore, 4 x 0.5 = 2 moles of potassium.

Mass of potassium = number of moles x molar mass

              Molar mass of potassium = 39g

Mass of potassium = 2 x 39 = 78g

5 0
3 years ago
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