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2 years ago

A solution is oversaturated with solute. Which could be done to decrease the oversaturation?

Physics

2 answers:

Gnoma [55]2 years ago

6 0

Dilute the solution, option a

salantis [7]2 years ago

3 0

Ans: Dilute the solution

Explanation:
To decrease the over-saturation, dilute the solution. Dilution is the process of decreasing the solute's concentration in the solution. It is usually done by mixing with more solvent. In other words, to dilute a solution means to add more solvent without the addition of more solute.

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The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is

Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

W = ΔEm

Where the work is

W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

Em₀ = K = ½ m v²

Final. When the system is at rest

$Em_{f}$ = $K_{e1}b +K_{e2}$ = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

∑ F = F - fr

Axis y

N- W = 0

N = W

The friction force has the equation

fr = μ N

fr = μ W

The job

W = (F – μ W) x

We substitute in the equation

(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values and solve

½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

x² 30 + 14.4 x - 1,125 = 0

x² + 0.48 x - 0.0375 = 0

We solve the second degree equation

x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

x = [-0.48 ± 0.617] / 2

x₁ = 0.0685 m

x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0

2 years ago

an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and

natulia [17]

Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

Time, t = 0.8 seconds

Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

$\Delta p=m(v-u)\\\\\Delta p=mv$

The change in momentum is equal to the impulse delivered. So,

$J=\Delta p=F\times t\\\\mv=F\times t\\\\v=\dfrac{Ft}{m}\\\\v=\dfrac{120\times 0.8}{55}\\\\v=1.745\ m/s$

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.

4 0

1 year ago

an input force of 50 Newtons is applied through a distance of 10 meters to machine with mechanical advantage of 3. If the work o

gladu [14]

The output of the machine is

(output work) = (output force) x (distance)

450 N-m = (output force) x (3 meters)

Divide each side
by 3 meters: Output force = (450 N-m) / (3 m)

= 150 newtons .

With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.

It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m). So everything
is nice and hunky-dory.
___________________________________

Well, my goodness !
I didn't even need to go through all of that.

Given:

-- The input force to the machine is 50 newtons.

-- The mechanical advantage of the machine is 3 .

That right there tells us that

-- The output force of the machine is 150 newtons.

We don't need any of the other given information.

5 0

2 years ago

A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

adelina 88 [10]

Answer:

x = 11.23 m

Explanation:

For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.

Let's reduce to SI system units

θ = 155 rev (2pi rad / rev) = 310π rad

α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²

Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)

w² = w₀² + 2 α θ

w =√ 2 α θ

w = √(2 4pi 310pi)

w = 156.45 rad / s

The relationship between angular and linear velocity

v = w r

v = 156.45 0.175

v = 27.38 m / s

In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

y = $v_{oy}$ t - ½ g t²

As it leaves the highest point its speed is horizontal

y = 0 - ½ g t²

t = √ (-2y / g)

t = √ (-2 (-0.820) /9.8)

t = 0.41 s

With this time we calculate the horizontal distance, because the constant horizontal speed

x = vox t

x = 27.38 0.41

x = 11.23 m

5 0

2 years ago

Two experiments are performed on an object to determine how much the object resists a change in its state of motion while at res

larisa86 [58]

Answer:

In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.

Explanation:

It is given that a student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.

In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass.

In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.

5 0

2 years ago