What is the force of gravity between you and a standard tennis ball if you are 1 meter away

How do rocks within Earth change as P waves pass? How do rocks within Earth change as P waves pass? Rocks within Earth both expa

ohaa [14]

Rocks within Earth both expand and contract as P waves pass

Explanation:

Rocks within the earth both expands and contracts as P-waves passes through them. P-waves are elastic waves.

Elastic waves behaves in such a way that they do not cause permanent deformation of rocks.
They can be said to cause elastic deformation when they travel through rocks.
They simply temporarily expand and contract the rock within a short period by causing the vibration of particles of the medium.
After a short while, the rock returns back to its original position as if nothing has happened to it.
These elastic waves are better called seismic waves.
P-waves are primary waves that can travel through any medium.

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6 0

3 years ago

A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee

Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

$s = ut +\frac{1}{2}at^{2}$

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

$s = \frac{1}{2}gt'^{2}$

$s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}$ (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

$s = v\times t''$

$s = 343\times t''$ (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t' (3)

Using eqn (2) and (3):

s = 343(3.5 - t') (4)

from eqn (1) and (4):

$4.9t'^{2} = 343(3.5 - t')$

$4.9t'^{2} + 343t' - 1200.5 = 0$

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

3 0

3 years ago

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

5 0

3 years ago

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How much tension must a rope withstand if it is used to accelerate 70.0 skier behind a boat at 1.50m/s^2?

SIZIF [17.4K]

It will be stand 46.67 all i did was divide both numbers but im not sure if im right but i hope i am hope i helped:)

6 0

3 years ago

A 2.0 kg particle moving along the z-axis experiences the

DochEvi [55]

At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.

Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.

If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Learn more here: brainly.com/question/20366032

6 0

3 years ago