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3 years ago

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Based on information from the graph above, what can be concluded about the relationship between the temperature of a solvent and

the solubility of a solute?
A. No conclusions can be drawn.

B. For most solutes, as temperature increases, the solubility is unaffected.

C. For most solutes, as temperature increases, the solubility increases.

D. For most solutes, as temperature increases, the solubility decreases.

1 answer:

Verdich [7]3 years ago

6 0

Pretty sure it’s C :)

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A point charge is placed 3m from a 4uc charge what is the strength of the electric field on the point charge at this distance ro

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The strength of the electric field on the point charge at this distance will be 4000 V/m.

<h3>What is the strength of the electric field?</h3>

The strength of the electric field is the ratio of electric force per unit charge.

The given data in the problem is;

Qis the unit charge = 4.0 × 10⁻⁶ C

E is the strength of the electric field

R is the distance from point charge = 3 m

The strength of the electric field is;

$\rm E = \frac{KQ}{R^2} \\\\ \rm E = \frac{9 \times 10^9 \times 4 \times 10^{-6} \ C}{3^2} \\\\ E= 4000 V/m$

Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.

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2 years ago

Infer how distance and speed in the motions of a clock parts are used to measure time

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A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee

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Quite low

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3 years ago

Read 2 more answers

Two cars, A and B , travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)

OLga [1]

A) Car A is initially ahead

B) The two cars are at the same point at the times: t = 0, t = 2.27 s and

t = 5.73 s

C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s

D) The two cars have same acceleration at t = 2.67 s

Explanation:

A)

The position of the two cars at time t is given by the following functions:

$x_A(t) = \alpha t + \beta t^2$

with

$\alpha = 2.60 m/s\\\beta = 1.20 m/s^2$

Substituting,

$x_A(t)=2.60t+1.20 t^2$

And

$x_B(t)=\gamma t^2 - \delta t^3$

with

$\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3$

Substituting,

$x_B(t)=2.80t^2-0.20t^3$

Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:

$x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m$

$x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m$

So, car A is initially ahead.

B)

The two cars are at the same point when their position is the same. Therefore, when

$x_A(t)=x_B(t)$

which means when

$2.60t+1.20t^2 = 2.80t^2-0.20t^3$

Re-arranging the equation, we find

$0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0$

One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation

$0.20t^2-1.6t+2.60=0$

which has two solutions:

t = 2.27 s

t = 5.73 s

So, these are the times at which the cars are at the same point.

C)

The distance between the two cars A and B is not changing when the velocities of the two cars is the same.

The velocity of car A is given by the derivative of the position of car A:

$v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t$

The velocity of car B is given by the derivative of the position of car B:

$v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2$

Therefore, the distance between the two cars is not changing when the two velocities are equal:

$v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0$

This is another second-order equation, which has two solutions:

t = 1.00 s

t = 4.33 s

D)

The acceleration of each car is given by the derivative of the velocity of the car A.

The acceleration of car A is:

$a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40$

While the acceleration of car B is:

$a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t$

So, the two cars have same acceleration when

$a_A(t)=a_B(t)$

And solving the equation, we find:

$2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s$

So, the two cars have same acceleration at t = 2.67 s.

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3 years ago

claculate the pressure exerted by the water on the bottom of a deep dam of 12m from its surface .(density of water =1000kg/m sqa

Rudiy27

Explanation:

We know that,

$hydrostatic \: pressure (p) = \alpha hg$

Where,

$\alpha = density \: of \: water \: = 1000$

h = Height at which pressure is to be calculated

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