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3 years ago

Describe how particles of solute behave when a solution is saturated.

Physics

1 answer:

Nadusha1986 [10]3 years ago

3 0

Explanation:

"When a solution reaches the point where it cannot dissolve any more solute it is considered "saturated." If a saturated solution loses some solvent, then solid crystals of the solute will start to form. This is what happens when water evaporates and salt crystals begin to form."

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A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

Ivan pushes a dresser with full drawers across his carpeted floor. What could Ivan do to reduce the amount of friction between t

aniked [119]

Answer:

Place some smooth tiles under the dresser

Smooth surfaces, like smooth tile, are easy to slide over. They create very little friction. Rough surfaces like carpet create much more friction.

remove the drawers from the dresser

Weight affects friction in that friction is directly proportional to the weight of the load one is moving. So reduce the weight, reduce the friction.

Explanation:

Speed does not impact friction, so moving the dresser slower won't help. Wind has nothing to do with the scenario, so that's not a correct option.

7 0

2 years ago

Read 2 more answers

A car of mass 1000 kg travelling at a velocity of 25 m/s collides with another car of mass 1500kg which is at rest. The two cars

Svetach [21]

Answer:

<em>The velocity of the two cars is 10 m/s after the collision.</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is

P=m.v

If we have a system of bodies, then the total momentum is the sum of them all

$P=m_1v_1+m_2v_2+...+m_nv_n$