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3 years ago

Maceo is making rock candy. Which best describes the steps she should take?

Physics

2 answers:

Ymorist [56]3 years ago

4 0

Answer:

<h2>Heat a saturated sugar water solution, dissolve more sugar, then let the solution cool</h2>

Explanation:

To make rock candy, you first need to place the water in a pan and bring it to boil. At the beginning you make a sugar water solution by adding a bit of sugar, that will ensure that you have the right temperature (remember, the more sugar you add, the harder is to dissolve it).

After you do the first process, then you continue to add the rest of the sugar while you stir, when it's completely dissolved, you remove the pan from the heat, and let the solution cool.

Therefore, the right answer is the first choice, because describes best the process.

Andru [333]3 years ago

3 0

Answer:

The answer is heat a saturated sugar water solution, dissolve more sugar, then let the solution cool

Explanation:

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An earthquake creates a type of wave that shakes the ground. If a large earthquake occurs in Greece, how can the waves be felt a

Cloud [144]

The sound waves travel through the sea which creates movement which can be felt all the way in Italy in the sea- like a domino effect

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3 years ago

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a

DENIUS [597]

Answer:

C = 4,174 10³ V / m^{3/4} , E = 7.19 10² / ∛x, E = 1.5 10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

V = C $x^{4/3}$

C = V / x^{4/3}

C = 220 / (11 10⁻²)^{4/3}

C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

V = E dx

E = dx / V

E = ∫ dx / C x^{4/3}

E = 1 / C x^{-1/3} / (- 1/3)

E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

E = 3 / C (0- (-1 / x^{1/3}))

E = 3 / 4,174 10³ (1 / x^{1/3})

E = 7.19 10² / ∛x

for x = 0.110 cm

E = 7.19 10² /∛0.11

E = 1.5 10³ N/C

6 0

3 years ago

A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee

Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

$s = ut +\frac{1}{2}at^{2}$

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

$s = \frac{1}{2}gt'^{2}$

$s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}$ (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

$s = v\times t''$

$s = 343\times t''$ (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t' (3)

Using eqn (2) and (3):

s = 343(3.5 - t') (4)

from eqn (1) and (4):

$4.9t'^{2} = 343(3.5 - t')$

$4.9t'^{2} + 343t' - 1200.5 = 0$

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

3 0

3 years ago

It takes 500 J of work to compress a spring 10 cm. What is the force constant of the spring? (b) When a 3.0 kg block is pushed a

NARA [144]

Answer:

Explanation:

a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²

500 = 5 x 10⁻³ k ,

k = (500/5) x 10³ = 10⁵ N/m

b )

k = 4.5 x 10¹ = 45 N/m

Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J

This energy gets dissipated by friction .

work done by friction = μ mg d

d is the distance traveled under friction

so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2

μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .

8 0

3 years ago

A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-

djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

$\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}$

$\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}$

$\frac{3-y}{y}=\sqrt{\frac{7}{2}}$

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0

3 years ago