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mina [271]
4 years ago
9

Limestone is an example of a chemical sedimentary rock. describe how limestone may form

Physics
1 answer:
Assoli18 [71]4 years ago
4 0
Limestone is formed (Chemically) thru calcium carbonate from lake or ocean.
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A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh
Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = \frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0
3 years ago
Horizontal abduction involves the movement of the arm or thigh in the transverse plane from a(n) _____ position to a lateral pos
ollegr [7]
There is a horizontal abduction and horizontal adduction in the transverse plane that is imaginary bisector that divides the body into two halves upper or top and lower or bottom halves.
<span>Horizontal abduction involves the movement of the arm or thigh from a medial or anterior position to a lateral position and in horizontal adduction, the movement is reverse that is from lateral position to anterior position.</span>
8 0
3 years ago
PLEASE HELP ME on this question.
VladimirAG [237]
The second one is the answer
4 0
3 years ago
A 14.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initially at rest on a horizontal surface. The clay s
ahrayia [7]

Answer:

86.53 m/s

Explanation:

Given:

Mass of clay (m) = 14.0 g = 0.014 kg

Mass of block (M) = 110 g = 0.110 kg

Initial speed of block (U) = 0 m/s

Sliding distance (d) = 7.50 m

Coefficient of friction between block and surface (μ) = 0.650

Let the initial speed of clay be 'u' and speed of clay and block just after collision be 'v'.

Now, momentum is conserved just before and just after collision.

Momentum just before collision = mu + 0 = mu

Momentum just after collision = (m + M)v

Therefore, mu=(M+m)v --------- (1)

Now, using newton's second law and we find the acceleration of the system.

The frictional force is given as:

f=\mu mg=-ma\\\\a=-\mu g

Now, using equation of motion, we can find the velocity just after collision.

0^2=v^2+2ad\\\\v=\sqrt{-2ad}\\\\v=\sqrt{-2\times (-\mu g)\times d}\\\\v=\sqrt{2\mu gd}

Plug in the given values and find 'v'. This gives,

v=\sqrt{2\times 0.650\times 9.8\times 7.50}\\\\v=\sqrt{95.55}=9.77\ m/s

Now, using equation (1) and substituting the given values, we get:

0.014u=(0.014+0.110)\times 9.77\\\\u=\frac{0.124\times 9.77}{0.014}\\\\u=86.53\ m/s

Therefore, the speed of the clay immediately before impact is 86.53 m/s.

7 0
3 years ago
Which of the planets do not have moons?
lubasha [3.4K]

Answer:

Mercury and Venus

Explanation:

7 0
4 years ago
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