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3 years ago

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How do we know that caramelizing sugar is a chemical change? A. It involves dissolving sugar in water. B. It changes the state o

f sugar from a solid to a liquid. C. It separates sugar from a sugar solution. D. It changes the sugar's color.

1 answer:

amid [387]3 years ago

8 0

Answer:

I think the correct answer is c

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PLEASSE HELP WILL MARK BRAINLIEST!!!!!!!!!!

faust18 [17]

They are halogen elements, or nonmetallic elements in the same GROUP, specifically group 17

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3 years ago

If mechanical energy is conserved, then a pendulum that has a potential energy of 20 J at its highest point and 0.5 J at its low

liraira [26]

At the highest point: kinetic energy is 0 due to the speed is 0

So the total mechanical energy is 20

Assume no frictions present, then the mechanical energy is conserved

So at the lowest point, kinetic energy = mechanical energy - potential energy

Answer will be 20 - 0.5 = 19.5 J

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4 years ago

Convert 5g/cm^3 into kg/m^3

Sliva [168]

Answer:

0.01135624

Explanation:

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3 years ago

Two astronauts of mass 100 kg are 2 m apart in outer space. What is the

fredd [130]

The force of gravity between the astronauts is $1.67\cdot 10^{-7}N$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

$m_1 = 100 kg\\m_2 = 100 kg$

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

$F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N$

Learn more about gravitational force:

brainly.com/question/1724648

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3 years ago

Read 2 more answers

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

Norma-Jean [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respecitvely; $a$ is the acceleration; and $\Delta x$ is the change in position.

So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

$\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}$

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0

3 years ago