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Oksanka [162]
2 years ago
12

Your teacher burns a piece of steel wool in class, demonstrating the chemical property, flammability. You are curious to see wha

t other things can demonstrate chemical properties. Which of these does not demonstrate a chemical change?
Physics
1 answer:
LekaFEV [45]2 years ago
3 0

Answer:

I assume by "which" of these you're looking for an example. Water freezing into ice or water, or evaporation, the process of turning from liquid into vapor, would not be chemical changes.

Explanation:

These are physical changes because they do not form a new substance, a chemical change requires a change in the chemical makeup of the substance.

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Suppose an X-ray binary is found in which the visible star is a 12 M⦿ red giant, the orbital period is 3.65 days, and the semima
nataly862011 [7]

Answer:

If the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

Explanation:

If a star's gravity is high enough, when it condenses on itself, it will form a black hole. Otherwise, it will create a large amount of highly dense matter, such as a neutron star. It can be said that if the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

8 0
2 years ago
You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel
Svetradugi [14.3K]

Answer:

a)693.821N/m

b)17.5g

Explanation:

We the Period T we can find the constant k,

That is

T = 2 \pi \sqrt{\frac{m}{k}}

squaring on both sides,

T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}

where,

M=hanging mass, m = spring mass,

k =spring constant

T =time period

a) So for the equation we can compare, that is,

y=T^2=0.0569x+0.0010

the hanging mass M is x here, so comparing the equation we know that

\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m

b) In order to find the mass of the spring we make similar process, so comparing,

\frac{4\pi^2}{k}m =0.001\\m=\frac{0.004k}{4\pi^2} =\frac{0.001*693.821}{4\pi^2}\\m=0.0175kg\\m=17.5g

3 0
2 years ago
2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initia
Elanso [62]

Answer:

Explanation:

The specific heat of gold is 129 J/kgC

It's melting point is 1336 K

It's Heat of fusion is 63000 J/kg

Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,

The first is E1 = 63000 J/kg x 1.5 = 94500 J

the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J

Therefore, all solid is not correct. You will have a mixture of solid and liquid.

For more detail, the difference between E1 and E2 is 7812 J, and that will melt

7812/63000 = 0.124 kg of the solid gold

8 0
2 years ago
the arm of a crane is 15.0 m long and makes an angle of 70.0 degrees with the horizontal. Assume that the maximum load for the c
Gnoma [55]

To find max torque you need component of force that is perpendicular to the direction of vector r. For second part just set the equation above equal to 2308 and solve for Fmax. Only difference is the angle

3 0
2 years ago
Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =
Semenov [28]
For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
r_1+r_2=L km
The second equation is going to the total force acting on the charge q3:
F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}
Let's solve this for r_1^2:
0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0
Now we have a quadratic equation with following parameter:
a=2\\ b=2L\\ c=-L^2
We know that two solutions are:
r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\
We need a positive solution. When we plug in all the numbers we get:
r_1=0.915\cdot 10^6$km

6 0
2 years ago
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