which criteria are responsible for deciding whether a heterogeneous mixture is a colloid or a suspension

1 answer:

6 0

I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. The criteria responsible for deciding whether a heterogeneous mixture is a colloid or a suspension is whether the particles remain suspended for an extended period of time. I hope it helps you.

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Spring stretches 14 cm when an object weighing 28 n is hung from it. what is the spring constant

daser333 [38]

The answer is 200 N/m

8 0

1 year ago

Armand hands the rock to Pierre, who is standing next to the trampoline. Explain how Armand moves, in terms of the forces acting

Nikolay [14]

Answer:

Armando's weight ,restored force created by the trampoline

a harmonic movement within the trampoline

Explanation:

In a trampoline we have two forces that actuate Armando's weight and the restored force created by the trampoline that depends on the deformation distance of the elastic canvas.

Amando's weight is vertical and directed towards the center of the Earth and has a constant value, this weight is balanced with the elastic force the springboard exerts on Armando in a vertical direction.

In general, when entering the trampoline, a small jump is made, this creates a speed that deforms the canvas until the speed is reduced to zero, at this point the elastic force is greater than the weight and the boy begins to climb, After the boy leaves the canvas he meets only the force of gravity and his speed decreases to zero and begins his fall.

In Summary Armando is in a harmonic movement within the trampoline

6 0

2 years ago

A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is c

bezimeni [28]

Answer:

The second dart leaves the gun two times as faster than the first one.

Explanation:

Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle

$U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}.$

Given an arbitrary $x$ and its double, $2x$ , launch velocities are

$v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}.$

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3 years ago

A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the

xxMikexx [17]

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

$tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2$