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3 years ago

9

How does Earth's surface and the structures on the surface change as a result of an earthquake? Help me pls and I will give Brai

nliest!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

EastWind [94]3 years ago

6 0

<span>Earthquakes often cause dramatic changes at Earth's surface. In addition to the ground movements, other surface effects include changes in the flow of groundwater, landslides, and mudflows. Earthquakes can do significant damage to buildings, bridges, pipelines, railways, embankments, dams, and other <span>structures</span></span>

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What Causes A Short Circuit?

slavikrds [6]

C. When there is too little resistance and the current increases

3 0

3 years ago

a force of 25 newtons moves a box a distance of 4 meeters in 5 seconds.the work done on the box is ? NM and the power is. ? nm/

miskamm [114]

Answer:

The work done on the box is 100 Nm

The power is 20 Nm/s

Explanation:

There is a force 25 newtons moves a box a distance of 4 meters in

5 seconds

The work done on the box is the product of the force and the distance

that the box moves ⇒ <em>work = force × distance</em>

The force = 25 newtons

the distance = 4 meters

Work = 25 × 4 = 100 NM

<em>The work done on the box is 100 Nm</em>

<em></em>

The force moves the box 4 meters in 5 seconds

The power is the rate of work

<em>The power = work ÷ time</em>

The work = 100 Nm

The time = 5 seconds

The power = 100 ÷ 5 = 20 Nm/s

<em>The power is 20 Nm/s</em>

6 0

3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

4 0

3 years ago

The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to

xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x Formula (1)

Where;

F is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21 inches

Problem development

We replace data in formula 1 to calculate K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

8 0

3 years ago

When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point

vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

$v = Velocity \rightarrow 20m/s$

$\lambda = Wavelength \rightarrow 35*10^{-2}m$

Therefore the frequency would be given as

$f = \frac{20}{35*10^{-2}}$

$f = 57.14Hz$

The frequency is directly proportional to the angular velocity therefore

$\omega = 2\pi f$

$\omega = 2\pi *57.14$

$\omega = 359.03rad/s$

Now the maximum speed from the simple harmonic movement is given by

$V_{max} = A\omega$

Where

A = Amplitude

Then replacing,

$V_{max} = (1*10^{-2})(359.03)$

$V_{max} = 3.59m/s$

Therefore the maximum speed of a point on the string is 3.59m/s

8 0

3 years ago