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3 years ago

Which of the following statements about gemstones is true? Choose one:

Physics

1 answer:

Ivan3 years ago

5 0

Answer:

Explanation:

Emerald is not a variety of corundum rather its a variety of beryl.

A precious or semiprecious stone may or may not be transparent.

The value of gemstone is derived mostly by its rarity.

Pegmatite are igneous rocks found underground near sheets of igneous rocks called batholiths. They are a source of gemstones made of topaz, tourmaline, beryl spodumene and spessartine.

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When an atom that has no charge loses two electrons, it becomes a

kicyunya [14]

Answer:

positive ion

Explanation:

due to the fact it loses electrons there is 2 more protons than electrons

3 0

3 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At <em>t</em> = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive <em>x</em>-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

2 years ago

A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured

Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}$

Put the value into the formula

$0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}$

$v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}$

$v_{3}=4.98\ m/s$

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

6 0

3 years ago

A man exerts a constant force to pull a 51-kg box across a floor at constant speed. He exerts this force by attaching a rope to

Vitek1552 [10]

Answer:

W=561.41 J

Explanation:

Given that

m = 51 kg

μk = 0.12

θ = 36.9∘

Lets F is the force applied by man

Given that block is moving at constant speed it mans that acceleration is zero.

Horizontal force = F cos θ

Vertical force = F sinθ

Friction force Fr= μk N

N + F sinθ = m g

N = m g - F sinθ

Fr = μk (m g - F sinθ)

For equilibrium

F cos θ = μk (m g - F sinθ)

F ( cos θ +μk sinθ) = μk (m g

Now by putting the values

F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10

F= 70.2 N

We know that Work

W= F cos θ .d

W= 70.2 x cos 36.9∘ x 10

W=561.41 J

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3 years ago

University Physics 9.68: Accelerating Compact Disc. A computer disc drive is turned on starting from rest and has constant angul

faust18 [17]

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3 years ago