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Nitella [24]
2 years ago
7

What kind of destructive force or forces will most likely change the way Stone Mountain looks over the next million years? A) vo

lcanoes B) landslides C) erosion and weathering D) earthquakes and faulting
Physics
2 answers:
zmey [24]2 years ago
8 0

D) earthquakes and faulting

zhannawk [14.2K]2 years ago
4 0

Answer:

D) Erosion and Weathering

Explanation:

Erosion forces rocks to crash together or crack apart, some rocks shatter and crumble, while others are worn away.

During erosion, exfoliation which is a type of weathering occurs in sheets along joints/ lines which runs throughout Stone Mountain. Once a sheet on the surface has been exfoliated and the sheet of rock beneath it is exposed the process begins again. This process gradually changes the way Stone Mountain looks.

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Earthquake information would most likely be shown on which type of map?
zmey [24]

Information about Earthquakes would normally be shown on a map called "geologic hazards"

6 0
3 years ago
If this atom has a balanced charged, how many protons would you expect to find in this atom?
avanturin [10]
The answer would be C because there is six electrons and so there will be six protons because the amount of protons and electrons have to be the same otherwise it would be an unbalanced particle and you wouldn't be able to touch the object it is in without worrying about something happening
3 0
3 years ago
Read 2 more answers
Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
kumpel [21]

Answer:

7800kg/m³

Explanation:

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

Given the density of iron = 7.8 g/cm3.

The SI units must be in kg/m³

7.8g = 7.8/1000 kg

7.8g = 0.0078kg

1cm³ = 0.000001m³

7.8g/cm³

= 0.0078/0.000001 kg/m³

= 7800kg/m³

Hence the density in SI unit is 7800kg/m³

4 0
3 years ago
D. What is the net force on the bowling ball rolling lane
3241004551 [841]

Answer:

Friction

Explanation:

3 0
2 years ago
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
3 years ago
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