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3 years ago

A series of processes on Earth’s surface and in the crust and mantle that slowly change rocks from one kind to another is called

1 answer:

6 0

A series of processes on Earth's surface and in the crust and in the mantle that slowly change rocks from one kind to another is called the rock cycle. This cycle is a group of changes. We all know that the rock has 3 kinds which are igneous rocks, the sedimentary rocks, and the metamorphic rocks. The igneous rocks in the rock cycle can change to the sedimentary rocks or to the metamorphic rocks. The sedimentary rocks can also change to the igneous rocks or to the metamorphic rocks. The metamorphic rocks can change to the igneous rocks or to the sedimentary rocks. All depends on the process it undergoes and the factors present to affect the change it undergoes.

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A baseball has a momentum of 6.0 kg m/s south and a mass of 0.15 kg. What is the baseball’s velocity? Show work

kupik [55]

Momentum of an object is defined as product of mass and velocity

so it is given as

$P = m* v$

here it is given that

mass = 0.15 kg

momentum = 6 kg m/s

so by plug in values in the above equation

$6 = 0.15 * v$

$v = \frac{6}{0.15}$

$v = 40 m/s$

so its speed will be 40 m/s

6 0

4 years ago

A 1.94-m-diameter lead sphere has a mass of 5681 kg. A dust particle rests on the surface. What is the ratio of the gravitationa

scoundrel [369]

To develop this problem we will apply Newton's laws regarding gravitational forces, both in space and on earth. From finding this relationship, leaving the variable of the dust mass open, we will find the relationship of the forces between the two surfaces. Our values are,

$\text{Diameter of the lead sphere} = D=1.940m$

$\text{Mass of the lead sphere} = m_1 = 5681kg$

$\text{Mass of the dust particle} = m_2$

Distance between the center of lead sphere to dust particle

$r = \frac{D}{2}$

$r = \frac{1.940m}{2}$

$r = 0.97m$

Gravitational force of the sphere on the dust particle:

$F = \frac{Gm_1m_2}{r^2}$

$F = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5681kg)(m_2)}{(0.97m)^2}$

$F = (4.027*10^{-7} N/kg)m_2$

Weight of the dust particle

$W = m_2 g$

$W = m_2 (9.8m/s^2)$

Ratio of F and W:

$\frac{F}{W} = \frac{(4.07*10^{-7}N/kg) m_2)}{m_2(9.8m/s^2)}$

$\frac{F}{W} = 4.153*10^{-8}$

Therefore the ratio is $4.153*10^{-8}$

3 0

3 years ago

CAN ANYONE HELP PLEASE I NEED THIS DONE BY TODAY!! I WILL MARK YOU BRIANLYIST.

bulgar [2K]

I have no idea my dude sorry

7 0

3 years ago

a satellite with a mass of 1800 kg is placed at an altitude of 594 km above the surface of Europa. What is the strength of the g

saul85 [17]

The formula for gravitational potential energy is Ep= mgh. Lets convert 594 km into m: 594 x 1,000 = 594,000 m. So we do: Ep= (1,800kg)(9.8N/kg)(594,000m) = (17,640N)(594,000) = 10,478,160,000 J.

4 0

3 years ago

1) what is a calorie?<br>

Bogdan [553]

I read it’s a unit of energy

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3 years ago