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3241004551 [841]
3 years ago
11

1. What ocean floor structures occur between 160 km and 1,050 km east of New Jersey? Between 2,000 km and 4,500 km? Between 5,30

0 km and 5,500 km? (10pts)
2. When a profile of a feature is drawn to scale, the horizontal and vertical scales must be the same. Does your profile give an accurate picture of the ocean floor? Explain. (20pts)

Physics
2 answers:
xenn [34]3 years ago
7 0
Hi 

Here is a chart to help you 

Hoped This Helps

netineya [11]3 years ago
6 0

Answer:

As explained below.

Explanation:

  • The structure if the ocean floor from <u>160 km to 1050 km east of new jersey is abyssal plains and hill, between the depth of 2000 km to 4500 km i.e 5100 meters deep lies the starting point of ocean deeps and deep-sea ridges, between the 5300 km to 5500 km i.e depth of 920 meters down lies the Continental slope  </u>
  • These are represented by the bathymetric profiles which provide a skyline view of hills and slopes, valleys and depression and .there horizontal and vertical ratios must be the same as 1:1.
  • However, they are generally so small that it's impossible to see them the profile sheet is drawn by manually selecting the VE = Horizontal Scale/Vertical Scale.  
  • As vertical exaggeration takes place the increasing features on a profile, hills appear to be higher, valleys deepen and the slopes among them get much steeper. Slopes, that, in reality,  are more gentle; steep slopes will appear to be stretched.
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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
3 years ago
Little Billie has a mass of 25 kg. The acceleration of gravity on the moon is one-sixth of the value on Earth. What is Little Bi
sineoko [7]

Answer:

Mass is constant everywhere,

But weight is different,

If earth g = 10 then moon's is 1.6666667

Now billie's weight in moon is 41.6667

6 0
3 years ago
What is the difference between a low tide and a high tide
Savatey [412]
A high tide means when the water has risen and is higher up(closer to high up land). Low tide is when it’s receded
6 0
2 years ago
Read 2 more answers
Why is the majority of Earth's freshwater not readily available for our use?
lara [203]
The answer is B. It is locked up in glaciers and ice caps.

Hope this helped. Good luck! Please give me a thanks and Brainliest.
8 0
3 years ago
Read 2 more answers
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
2 years ago
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