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2 years ago

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The chart below lists the densities of various gemstones. If a gemstone has a mass of 6.24 g and a volume of 1.98 cm cubed, whic

h gemstone is it
A. Opal
B. Diamond
C. Garnet
D. Topaz

1 answer:

xxMikexx [17]2 years ago

5 0

Answer:

C. Garnet

Explanation:

density = mass/volume

density = 6.24g/1.98cm^3

density = 3.15 g/cm^3 (Garnet)

Hope this helps!

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Gravity pushes objects away from each other TRUE OR FALSE

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Answer:

false

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2 years ago

Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (

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Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

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4 kg (2.5, 0)

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As we know that the formula of center of gravity is given as

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$0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}$

$10 + 8.3 x = 0$

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Similarly for y direction we have

$y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}$

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3 years ago

Un soldado de 1.80 m de altura ha sido herido por una bala, cuya masa es de 100 g. Deciden colocar al soldado en una centrifugad

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3 years ago

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Answer:

a

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The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

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Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

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=> $a_c = gtan \theta$

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substituting value

$T_x = 0.344 * 0.9574$

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The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

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3 years ago

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Answer:

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