What is the relationship between karst development and topography?

1 answer:

5 0

The karst topography is typically defined as a geographic location characterized by a rugged terrain containing landscapes like underground rivers, fissures, and cracks. It is mainly due to the dissolution of the bedrock due to a much heavier precipitation taking place in the geographic location.

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Ball B, with a mass of 30kg, is moving to the left at 10 m/s. With what velocity should ball A, with a mass of 10kg, move to the

bogdanovich [222]

Using the conservation of momentumwhere momentum = mass x velocity
so the let x be the initial velocity of the ball B

(30 kg) (-10 m/s) + (10 kg) (x) = (30 kg) (30 m/s) + ( 10 kg) (-10 m/s)
-300 + 10 x = 900 - 10010x = 900 - 100 + 30010x = 1100x = 1100 / 10x = 110 m/s

6 0

3 years ago

An object will maintain a constant speed until a force acts on it true or false

Scorpion4ik [409]

You do not require a force to keep something moving. You only require a force to get it moving. Or to stop it moving. In your everyday experiences, something you get moving seems to come to a stop after you stop pushing it. It is because there are forces (friction) that make it stop. Without those forces, the object would just keep moving. So this would mean the answer would be True.

3 0

3 years ago

Car A with a mass of 725 kilograms is traveling east at an initial velocity of 15 meters/second. It collides head–on with car B,

ikadub [295]

Answer:

$p_t_o_t_a_l=250kg\frac{m}{s}$

Explanation:

The total momentum of a system is defined by:

$(mv)_t_o_t=m_1v_1+m_2v_2+...$

Where,

$(mv)_t_o_t$ is the total momentum or it could be expressed also as $p_t_o_t_a_l$ .

$m_1$ and $m_2$ represents the masses of the objects interacting in the system.

$v_1$ and $v_2$ are the velocities of the objects of the system.

Remember: The momentum is a fundamental physical magnitude of vector type.

We have:

$m_1=725 kg$

$v_1=15\frac{m}{s}\\m_2=625 kg$

We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be negative. It goes in a different direction from car A.

$v_2=-17\frac{m}{s}$

Then the total momentum of the system is:

$p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}$