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Mars2501 [29]
3 years ago
5

The breakdown of rocks by chemical action of air and water is called

Physics
2 answers:
Dovator [93]3 years ago
8 0

Answer:

Chemical weathering

Explanation:

This changes the molecular structure of rocks and soil. For instance, carbon dioxide from the air or soil sometimes combines with water in a process called carbonation. This produces a weak acid, called carbonic acid, that can dissolve rock.

Hunter-Best [27]3 years ago
3 0

Answer:

Chemical weathering

Explanation:

This changes the molecular structure of rocks and soil. For instance, carbon dioxide from the air or soil sometimes combines with water in a process called carbonation. This produces a weak acid, called carbonic acid, that can dissolve rock.

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Objects in free fall are weightless. true or false???
frez [133]

The answer is false

7 0
3 years ago
What does the potential energy and kinetic energy of an object depend on ?
alexdok [17]

Answer:

Both, potential energy and kinetic energy depends on mass. The higher the mass, the higher the energy. However, the difference is that potential energy depends on vertical height whereas kinetic energy depends on the velocity.

Explanation:

From the formula we can see that;

Potential Energy = mass* gravitational acceleration *vertical height.

Kinetic Energy = 0.5 * mass * (velocity)^2

4 0
3 years ago
A train travels 90 kilometers in 2 hours, and then 66 kilometers in 2 hours. What is its average speed?
Inessa05 [86]
90+66=156
156/2=78

Reply:78kilometers in 2 hours.
3 0
3 years ago
Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
andreev551 [17]

Answer:

the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = m_A

then, the mass of object B = m_B = \frac{m_A}{4}

work done on object A = -18 J

work done on object B = -18 J

let v_i be the initial speed

let v_f be the final speed

For object A;

K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2  = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ -  v_i^2 )\ =- 18\\\\v_f^2 \ -  v_i^2  = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ -  v_i^2  = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\

v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\

Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B =  \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\

\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i

Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

3 0
2 years ago
How does temperature influence a chemical change?
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Your weight #sorryfortheweight
8 0
3 years ago
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