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Zarrin [17]
2 years ago
13

I NEED A SCIENCE EXPERT TO GIVE ME THE RIGHT ANSWER TO THESE ASAP NO LINKS !!!

Physics
2 answers:
alexdok [17]2 years ago
8 0

Answer:

it IS the control variable

Explanation:

ddd [48]2 years ago
6 0

Answer:

1 st question: Control variable

2nd question: random variable

3rd question: if the two objects are dropped from the same height they will hit the ground at the same time

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Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was doubled, and
Lady_Fox [76]

Answer:

  • Distance, such that displacement over time is taken into account doesn't change regardless of the length in x meters or mass in kg
  • So the force will remain constant given 16 units
6 0
3 years ago
This formula equation is unbalanced.
Grace [21]
The answer is:  [C]:  "4" .
___________________________________________________
Note:  To balance this equation, the coefficient, "4", should be placed in front of the PCl₃ ;  and the coefficient, "6", should be placed in front of the Cl₂ .
________________________________________________________
The balanced equation is:
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             P₄ (s) + 6 Cl₂ (g)  <span>→ 4 </span>PCl₃ (l)   .
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6 0
3 years ago
Why is it sometimes difficult for scientists to identify a a fold or fault?
Sauron [17]

Folds and faults are difficult to identify because they occur in the interior of rocks and also due to the dense nature of the materials.

<h3>What are faults and folds?</h3>

Faults are lines of weakness are present in materials dues to uneven positioning of the particles of the material.

Folds occurs when infolds occur in materials.

Faults and folds usually occur in rocks.

Folds and faults are difficult to identify because they occur internally and also due to the dense nature of the materials.

Learn more about faults and folds at: brainly.com/question/14240712

4 0
2 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
Which value is equivalent to 178 centimeters?
tester [92]

178 centimeters is exactly, precisely, the same length as ...

-- 0.00178 kilometer
-- 1.78 meters
-- 1,780 millimeters
-- 1,780,000 micrometers
-- 1,780,000,000 nanometers

7 0
3 years ago
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