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Juli2301 [7.4K]
3 years ago
9

If a layer was deposited but does not appear in the rock record, what has occured

Physics
2 answers:
jekas [21]3 years ago
7 0
Erosion. Erosion<span> is the action and movement of a surface process (e.g. </span>water or wind<span>) that removes and transports a material layer</span><span> from one location</span><span> to another. Once eroded, the deposited layer will not appear in the rock record. </span>
nalin [4]3 years ago
5 0
I think you forgot to give the choices along with the question. I am answering the question based on my research and knowledge. <span>If a layer was deposited but does not appear in the rock record, the thing that happened is erosion. I hope that this is the ans wer that has actually come to your desired help.</span>
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A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
3 years ago
An action potential arriving at the presynaptic terminal causes what to occur?
Ulleksa [173]

Answer:

Voltage-gated calcium ion channels open, and calcium ions diffuse into the cell

5 0
3 years ago
A person wants to fire a water balloon cannon such that it hits a target 100m100m away. if the cannon can only be launched at 45
vladimir2022 [97]
<span>31.3 m/s Since the water balloon is being launched at a 45 degree angle, the horizontal and vertical speeds will be identical. Also the time the balloon takes to reach its peak altitude will match the time it takes to fall. So let's create a few expressions about what we know. Distance the water balloon travels at velocity v for time t d = vt Total time required for the entire trip is double since the balloon goes up, then goes down t = 2v/a Now let's plug in the numbers we have, assuming the acceleration due to gravity is 9.8 m/s^2 t = 2v/9.8 100 = vt Substitute 2v/9.8 for t in the 2nd formula 100 = v(2v/9.8) Solve for v. 100 = v(2v/9.8) 100 = 2v^2/9.8 980. = 2v^2 490 = v^2 22.13594 = v So we now know that both the horizontal velocity and vertical velocity needed is 22.13594 m/s. Let's verify that 2*22.13594 / 9.8 = 4.51754 So it will take 4.51754 second for the balloon to hit the ground after being launched. 4.51754 * 22.13594 = 100 And during that time it will travel 100 meters horizontally. But we need to know the total velocity. And the Pythagorean theorem comes to the rescue. Just square the 2 velocities, add them together, and take the square root. We already know the square is 490 from the work above, so sqrt(490+490) = sqrt(980) = 31.30495 m/s</span>
3 0
3 years ago
Two particles A and B start simultaneously from a Point P with velocities 20 m/s and 30 m/s respectively. A and B move with acce
zysi [14]

Answer:

<u>20 m/s</u>

Explanation:

<u>Given</u>

  • u(A) = 20 m/s
  • u(B) = 30 m/s
  • acceleration equal in magnitude but opposite in direction

<u>Solving</u>

  • Velocity of A at Q = 30 m/s
  • From, P to Q, <u>Δv(A) = 30 - 20 = +10 m/s</u>
  • Therefore, velocity of B at Q will be decreased by 10 as it is equal in magnitude but opposite in direction to A
  • Δv(B) = v(B at Q) - u(B at P)
  • -10 m/s = v(B at Q) - 30 m/s
  • v(B at Q) = 30 - 10 = <u>20 m/s</u>
6 0
3 years ago
A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r
marusya05 [52]

As we know that here no air resistance while ball is moving in air

So here we will say that

initial total energy = final total energy

KE_i + U_i = KE_f + U_f

here we know that

Ui = U_f = 0 (as it will be on ground at initial and final position)

so we will say

KE_i = KE_f

since mass is always conserved

so we will say that final speed of the ball must be equal to the initial speed of the ball

so we have

v_f = v_i = 30 m/s

6 0
3 years ago
Read 2 more answers
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