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3 years ago

Quartz, gold and calcite are examples

Physics

2 answers:

Vesnalui [34]3 years ago

7 0

They are examples of minerals.

Ymorist [56]3 years ago

4 0

Quartz, gold and calcite are all good examples of minerals. They all fit to the characteristics of minerals.
Minerals are usually inorganic and in a solid state. It has crystal structure and can be represented by a chemical formula. Minerals and rocks are distinct. It can be described in many different physical properties in relation to their chemical composition and structure. The most common characteristics of minerals includes, colour, streak, lustre, habit, hardness, tenacity, cleavage, fracture, specific gravity and parting..

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Major storms get their energy from WATER VAPOR in the air. Which step of the water cycle would give energy to a storm

IrinaK [193]

Evaporation is the process of water being evaporated and rising. This water vapor condenses at a certain point and mixes with aerosols to create a cloud.

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3 years ago

If a tomato is considered a fruit... then is ketchup considered a smoothie?

Anna71 [15]

Answer:

depends... do you add suger to your ketchup?

Explanation:

8 0

3 years ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

Given:

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

Assume:

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago

In a science museum, a 140 kg brass pendulum bob swings at the end of a 16.8 m -long wire.

Mice21 [21]

The period of the pendulum is 8.2 s

Explanation:

The period of a simple pendulum is given by the equation:

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

T is the period

We notice that the period of a pendulum does not depend at all on its mass, but only on its length.

For the pendulum in this problem, we have

L = 16.8 m

and

$g=9.8 m/s^2$ (acceleration of gravity)

Therefore the period of this pendulum is

$T=2\pi \sqrt{\frac{16.8}{9.8}}=8.2 s$

#LearnWithBrainly

3 0

3 years ago

On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera

pav-90 [236]

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s

8 0

4 years ago