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algol13
3 years ago
11

Help me please Ima give brainiest

Physics
2 answers:
julsineya [31]3 years ago
4 0
Sand. You can identify tiny sugar crystals.


creativ13 [48]3 years ago
3 0

<em>Answer: </em><em>vinegar </em>

A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture. ... A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead, form two separate layers. Each of the layers is called a phase.

<em>Hope it helps...</em>

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Pyrite is called "fool's gold" because it's looks a lot like gold. Which properties can be used to tell gold and pyrite apart?
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Density, streak, and geochemical signature through use of inductively coupled plasma mass spectrometry or Fourier transform infrared spectroscopy
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3 years ago
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How does moisture affect a sponge
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<span>the amount of water a sponge can absorb depends on many factors. The material the sponge is made out of, the structure of the sponge and the size of the sponge all affect how much water the sponge can absorb.</span>
6 0
3 years ago
A 300 g block connected to a light spring with a force constant of k = 3 N/m is free to oscillate on a horizontal, frictionless
Usimov [2.4K]

Answer:

Time period, T = 1.98 seconds

Explanation:

It is given that,

Mass of the block, m = 300 g = 0.3 kg

Force constant of the spring, k = 3 N/m

Displacement in the block, x = 3 cm

Let T is the period of the motion of the block. The time period of the block is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

T=2\pi \sqrt{\dfrac{0.3\ kg}{3\ N/m}}    

T = 1.98 seconds

So, the period of the motion of the block is 1.98 seconds. Hence, this is the required solution.

7 0
3 years ago
A 11kg block slides up a 30° inclined plane at a constant velocity. The coefficient of friction between the block and the plane
astra-53 [7]

Answer:

The magnitude of the applied force is 94.74 N

Explanation:

Mass of the block, m = 11 kg

Angle of inclination of the plane, \theta = 30^{\circ}

Friction coefficient, \mu_{k} = 0.2

Now,

Normal force that acts on the block is given by:

F_{N} = mgcos\theta + Fsin\theta           (1)

Now, to maintain the equilibrium parallel to ramp the forces must be balanced.

Thus

Fcos\theta = \mu_{k}F_{N}                       (2)

From eqn (1) and (2)

Fcos\theta = \mu_{k}(mgcos\theta + Fsin\theta)

F(cos\theta - \mu_{k}sin\theta) = \mu_{k}mgcos\theta

F = \frac{\mu_{k}mgcos\theta}{cos\theta - \mu_{k}sin\theta}

F = \frac{0.2\times 11\times 9.8cos30^{\circ}}{cos30^{\circ} - 0.2\times sin30^{\circ}}

F = 94.74 N

3 0
4 years ago
The Enterprise goes into orbit around a mysterious planet. The ship moves at 4200 m/s in a circle of radius 4.91 x 10^7 m. What
NeX [460]

Answer:

M = 1.3*10^25 Kg

Explanation:

3 0
3 years ago
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