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Lana71 [14]
3 years ago
12

The majority of earthquakes worldwide occur at all but one location. That is A)at tectonic plate boundaries.B)where sediments de

posit and form new layers.C)where plates stick and then jump past each other.D)where an ocean plate slides under a continental plate.
Physics
1 answer:
DENIUS [597]3 years ago
6 0

Answer:b

Explanation:

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Si aplicas una fuerza constante (F=constante) a un cuerpo de masa m1, dicho cuerpo se acelerará a1. Nuevamente, si aplicas la mi
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<span>Debido a que la fuerza es constante, a medida que la masa aumenta, la aceleración disminuye. Por lo tanto, la mayor aceleración sería a1.</span><span>Utilicé Google traductor, lol. Sin embargo, espero que esto ayude.</span>
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Hurricanes along the Gulf Coastal areas of Texas increase the rate of coastal erosion.
4vir4ik [10]

Answer:

its the last question

Explanation:

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3 0
3 years ago
A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
4 years ago
U guys are the best I’m surprised I’m getting all of this right
ella [17]
The answer is B, 1/4 of the cake
5 0
3 years ago
Read 2 more answers
Spin dynamics during chirped pulses: applications to homonuclear decoupling and broadband excitation
Likurg_2 [28]

Swept-frequency pulses have found use in a variety of fields, including spectroscopic methods where effective spin control is necessary.

To find more, we have to study about the spectroscopic methods.

<h3>What is homonuclear decoupling and broadband excitation?</h3>
  • A thorough understanding of the evolution of spin systems during these pulses is crucial for many of these applications since it not only helps to describe how procedures work but also makes new methodologies possible.
  • Broadband inversion, refocusing, and excitation employing these pulses are some of the most popular applications in NMR, ESR, MRI, and in vivo MRS in magnetic resonance spectroscopy.
  • A generic expression for chirped pulses will be presented in this study, along with numerical methods for calculating the spin dynamics during chirped pulses using solutions along with extensive examples.

Thus, we can conclude that, the swept-frequency pulses have found use in a variety of fields, including spectroscopic methods where effective spin control is necessary.

Learn more about the broadband excitation here:

brainly.com/question/19204110

#SPJ4

7 0
2 years ago
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