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4 years ago

when liquid gold becomes a solid, it's particles get closer together. which best describes what happened to the gold?

Physics

2 answers:

Pepsi [2]4 years ago

5 0

I believe it’s option D. The partials get closer together

Hope this helps :)

Sergio [31]4 years ago

5 0

Answer:

D is correct answer trust me I got 100% on Edgenuity 2020

Explanation:

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Which statement describes a physical property related to the particles that make up a material? A. Gold ions in a piece of jewel

ohaa [14]

Answer: Metallic bonds between tin ions cause a sheet of tin to be malleable

Explanation:

4 0

3 years ago

Read 2 more answers

A roller coaster starts from rest at point A. What is its speed at point C if the track is frictionless.

OleMash [197]

At A, coaster is only associated with potential energy.
At B, coaster is associated with kinetic as well as potential energy.
Since the track is frictionless, no energy will be lost when coaster reaches from point A to point B. Therefore, according to conservation of energy, total energy at A should be equal to total energy at B.
Total energy at A = mgh = mg(12)
Total energy at B = mgh+ mv²/2 = mg(2) + mv²/2
∴12mg = 2mg + mv²/2
∴(12g-2g)×2 = v²
∴v² = 20g
∴v = 14m/s.

Again conserving energy at points B and C.
Total energy at B = 2mg + m(14)²/2
Total energy at C = 4mg + mv²/2
∴2mg + m(14²)/2 = 4mg + mv²/2
Solving this you get,
v = 12.52 m/s.
Therefore, speed of roller coaster at point C is 12.52 m/s.

8 0

4 years ago

The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i

Archy [21]

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

$T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1$

Block 2

$T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2$

Block 3

$- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3$

Solving Eq1,2,3 simultaneously

Divide 1 and 2

$\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4$

Put Eq 4 into Eq3

$T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5$

Put Eq 5 into Eq2 and solve for a

$a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6$

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

$T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\$

5 0

4 years ago

A graduated cylinder contains 62 ml of water. When a small metal block is added to the water, the volume of the water increases

lawyer [7]

Answer:

The density of the block is 7.4g/ml.

Explanation:

We can determine the volume of the metal block by taking the difference between the volumes measured in the graduated cylinder:

$V_{block}=65.5ml-62ml\\\\V_{block}=3.5ml$

Now, as we know that the average density of an object is calculated dividing its mass by its volume, we can calculate the density ρ of the metal block using the expression:

$\rho_{block}=\frac{m_{block}}{V_{block}}\\\\\rho_{block}=\frac{26g}{3.5ml}\\\\\rho_{block}=7.4\frac{g}{ml}$