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3 years ago

What is rock made of?

2 answers:

6 0

Almost all rocks made of minerals, but different rocks contain different mixtures of minerals. Granite, for example, consists of quartz, feldspar, and mica.

lesya [120]3 years ago

4 0

Answer:

They are made from Minerals.

Explanation:

Rocks are composed of grains of minerals, which are homogeneous solids formed from a chemical compound arranged in a orderly manner. The aggregate minerals forming the rock are held together by chemical bonds. the types and abundance of minerals in a rock are determined by the manner in which it was formed.

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A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o

Ierofanga [76]

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: 

Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is:

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx.

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have:

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx.

Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)

W = ∫ 1250π x dx
W = 1250π ∫ x dx
W = 625π * (x2 – x1)

x2 = 14 ft + 15 ft = 29 ft

x1 = 14 ft + 1 ft = 15 ft

W = 625π * (29^2 - 15^2)
W = 385,000π ft-lbs = 1,209,513.17 ft-lbs

3 0

4 years ago

A 1 170.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 800.0 kg t

Stella [2.4K]

Answer:

Explanation:

mass of car, m1 = 1170 kg

mass of truck, m2 = 9800 kg

initial velocity of car, u1 = 25 m/s

initial velocity of truck, u2 = 20 m/s

final velocity of car, v1 = 18 m/s

Let the final velocity of truck is v2.

(a) Use the conservation of momentum

m1 x u1 + m2 x u2 + m1 x v1 + m2 x v2

1170 x 25 + 9800 x 20 = 1170 x 18 + 9800 x v2

29250 + 196000 = 21060 + 9800 v2

v2 = 20.84 m/s

(b) Total kinetic energy before collision,

Ki = 0.5 x 1170 x 25 x 25 + 0.5 x 9800 x 20 x 20

Ki = 365625 + 1960000 = 2325625 J

Total final kinetic energy

Kf = 0.5 x 1170 x 18 x 18 + 0.5 x 9800 x 20.84 x 20.84

Kf = 189540 + 2128097.44

Kf = 2317637.44 J

Change in energy = 2325625 - 2317637.44 = 7987.56 J

4 0

4 years ago

Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water le

mafiozo [28]

Answer:

v = 31.84 cm/s or 0.318 m/s

the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s

Explanation:

Given;

Diameter of hose d = 2.76 cm

Volume filled V = 20.0 L = 20,000 cm^3

Time t = 1.45 min = 105 seconds

The volumetric flow rate of water is;

F = V/t = 20,000cm^3 ÷ 105 seconds

F = 190.48 cm^3/s

The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.

F = Av

v = F/A

Area A = πd^2/4

Speed v = F/(πd^2/4)

v = 4F/πd^2 ......1

Substituting the given values;

v = (4×190.48)/(π×2.76^2)

v = 31.83767439628 cm/s

v = 31.84 cm/s or 0.318 m/s

the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s

3 0

3 years ago

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}$

$E=2.41\times10^{3}\ V/m$

Hence, The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

3 0

4 years ago

How would I have to calculate this?

guajiro [1.7K]

Answer:

by the product of mass and acceleration..

3 0

2 years ago