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3 years ago

8

you mix two ionic Solutions after you mix them in a beaker the solution turns cloudy after 5 minutes the solution becomes clear

and a white powder has settled on the bottom of your maker is it physical or chemical change

1 answer:

shepuryov [24]3 years ago

4 0

It's a chemical change. When mixed, the ion exchange occurs. Copper and sodium switch places in the compounds.

You might be interested in

Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of

alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is <u>29.89 mi/h</u> which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

$\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx 87.084 \ ft./s$

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s

≈ <u>87.084 ft./s</u>

<u /> $v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h$ <u />

The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

$v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}$

$v = \dfrac{40,000 \times 30 + 150 \times 0}{40,000 + 150} \approx 29.89$

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>

Learn more here:

brainly.com/question/18326789

3 0

2 years ago

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2 h, and Car B traveled the dista

lions [1.4K]

The answer is 60 mph.

The speed (v) is distance (d) per time (t): v = d/t

Car A:
v1 = ?
t1 = 2 h
d1 = ?
___
v1 = d1/t1
d1 = v1 * t1

Car B:
v2 = ?
t2 = 1.5 h
d2 = ?
___
v2 = d2/t2
d2 = v2 * t2

<span>Two cars traveled equal distances:
d1 = d2
</span>v1 * t1 = v2 * t2

<span>Car B traveled 15 mph faster than Car A:
v2 = v1 + 15

</span>v1 * t1 = v2 * t2
v2 = v1 + 15
________
v1 * 2 = (v1 + 15) * 1.5
2v1 = 1.5v1 + 22.5
2v1 - 1.5v1 = 22.5
0.5v1 = 22.5
v1 = 22.5/0.5
v1 = 45 mph

v2 = v1 + 15
v2 = 45 + 15
v2 = 60 mph

8 0

3 years ago

ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po

Levart [38]

Answer:

$6.046N$

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

$F_n_e_t=F_g+F_t$

$F_t$ is the tension force. $F_g=9.8N/kg$ is the force of gravity.

$F_n_e_t=ma_c=mv^2/r\\$

where $r$ is the rope's radius from the fixed point.

From the net force equation above:

$F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N$

Hence the tension force is 6.046N

8 0

3 years ago

An object that is dropped straight down from a height of 100 m has a vertical change in position that is less than that of an id

sergey [27]

Objects dropped straight or thrown horizontally from the same height
change their vertical velocity at the same rate, and fall through equal
vertical distances in equal time intervals.

The statement is false.

5 0

3 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago