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3 years ago

Which type of wave affects the surface of the land by causing it to rise and fall like waves on an ocean?

1 answer:

5 0

It is a seismic wave that effects the surface of the land by causing it to rise and fall like waves in an ocean. Seismic waves can be broken down into a number of different types. Specifically, it is a Raleigh surface wave that behaves much like water waves since it moves the ground up and down as it travels. Seismic waves are produced in earthquakes.

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A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and ampl

scoundrel [369]

Answer:

Explanation:wave=wavelength×frequency,

3 0

3 years ago

An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t

docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

$y_1(t)=h+u_1 t + \frac{1}{2}gt^2$

where:

h = 50 m is the initial height

$u_1=0$ is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

$g=-9.8 m/s^2$ is the acceleration of gravity

So, its vertical position can be rewritten as

$y_1(t)=50-4.9t^2$

The position of object 2 instead can be written as

$y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2$

where

$u_2 sin \theta$ is the initial vertical velocity, where

$u_2 = 50 m/s$ is the initial velocity

$\theta=30^{\circ}$ is the angle of projection

Substituting, we get:

$y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2$

The two objects collide when their vertical position is the same, so:

$y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2$

And solving for t, we find:

$50=25t\\t= 2 s$

Note that this means that the two object at t = 2 s have the same vertical position: however, this is not true for the horizontal position.

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

$y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m$

We can verify that at the same time, the vertical position of object 1 is the same:

$y_1(t)=50-4.9t^2=50-4.9(2.0)^2=30.4 m$

This means that the two objects have the same vertical position at 30.4 m.

However, in reality, the two objects do not collide. In fact, object 1 is moving in the horizontal direction with constant velocity

$v_{1x}=25 m/s$

So its horizontal position at t = 2.0 s is

$x_1(2.0)=v_{1x}t=(25)(2.0)=50 m$

While object 2 is moving in the horizontal plane with velocity

$v_{2x}=u_2 cos \theta=(50)(cos 30^{\circ})=43.3 m/s$

So its horizontal position at t = 2.0 s is

$x_2(2.0)=v_{2x}t=(43.3)(2.0)=86.6 m$

So in reality, the two objects do not collide, if they start from the same x-position.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

Two blocks travel along a level frictionless surface. Block A is initially moving to the right, while block B is initially at re

WARRIOR [948]

Answer:

Explanation:

The magnitude of the change in momentum of block A is <u>equal and opposite </u>the magnitude of the change in momentum of block B.

According to law of conservation of momentum , there should not be any change in the momentum in the system .The system consists of A and B . So if some change in momentum has occurred in A , the other must have equal and opposite change in momentum so that they get cancel out and net change in momentum of system is zero.

4 0

3 years ago

A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas

Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

$\varepsilon' =\dfrac{S_y}{E}$

$\varepsilon' =\dfrac{240}{110\times 1000}$

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

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3 years ago

which statement correctly describes the relationship between a black body radioterapia it’s tmeptaute and the electromagnic radi

jasenka [17]

It’s A if there’s multiple choices

7 0

3 years ago

Read 2 more answers