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UkoKoshka [18]
3 years ago
15

How does weathering, erosion and deposition shape the Earth and contribute to the rock cycle? Weathering, erosion and deposition

shape the Earth by ______ and contribute to the rock cycle by ______.
Physics
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

Weathering, erosion and deposition over millions of years can destroy and break apart rocks which in turn can cause sediments which is directly involved in the rock cycle.

Explanation:

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considere que o calor específico de um material presente nas cinzas seja c=0,8j/gc. Supondo que esse material entre na turbina a
drek231 [11]

Answer:

3120J

Explanation:

Given parameters:

C  = Specific heat capacity  = 0.8J/g°C

Initial temperature  = 20°C

Mass given   = 5g

Final temperature  = 800°C

Unknown:

Energy given to the mass  = ?

Solution:

To find the energy given to the mass, let us simply use the expression below:

          H   =   m   c   ΔT

H is the unknown, the energy supplied

m is the mass of the substance

c is the specific heat capacity

ΔT is the change in temperature

Input the variables;

            H    = 5  x   0.8    x    (800 - 20)  = 3120J

7 0
3 years ago
The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 f
Andrej [43]

Answer:

35.14°C

Explanation:

The equation for linear thermal expansion is \Delta L = \alpha L_0\Delta T, which means that a bar of length L_0 with a thermal expansion coefficient \alpha under a temperature variation \Delta T will experiment a length variation \Delta L.

We have then \Delta L = 0.481 foot, L_0 = 1671 feet and \alpha = 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a \Delta T =\frac{\Delta L }{\alpha L_0} = 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.

Since our temperature on a cool spring day was 13.0°C, our new temperature must be T_f=T_0+\Delta T = 35.14°C

3 0
3 years ago
As the average kinetic energy of a substance increases, the tempature of the sample
taurus [48]
I have no clue. Sorry bro
6 0
2 years ago
Read 2 more answers
A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070
likoan [24]

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

8 0
3 years ago
0. 85 kg of lead, specific heat 128 J/kgC, is heated from 50 C to 95 C. How much heat energy did the sample absorb?
Zanzabum
4896

0.85 x 45 x 128 = 4896

Change in energy = specific heat capacity x mass x change in temperature
5 0
2 years ago
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