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nignag [31]
3 years ago
10

Why did most scientist reject wegeners theory for nearly a half century

Physics
2 answers:
zlopas [31]3 years ago
7 0
He couldn't provide as much satisfactory explaining for the force that pushes or pulls the continents also because it contradicted their ideas about how mountains form.
SOVA2 [1]3 years ago
6 0
I think they were also too skeptic to believe the ocean did move or pull apart, even today do you believe that the continents broke from one big flat plate, and that they pulled apart? They also wonder what large force would be responsible for the movement.

It was much later that evidences from plant and animal features that had similarity from two different planets came up that scientists began accepting the idea of continental drift.
And similar rock strata from two different opposite continents, showed similar rock strata.
All these evidences came up much later after Alfred Wengener.

So Alfred Wengener was honored Posthumously.
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Matthew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the ba
MakcuM [25]
At the very top of the peak and on the way down
7 0
3 years ago
Read 2 more answers
If you were looking for a metalloid on the periodic table,the best place to look would be?
vovikov84 [41]

Answer:

Long the step line

Explanation:

8 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
X<br> x<br> 2.5 N<br> 1.7 N<br> 2.5 N
Olin [163]

Answer:

2.5*1.7/2.5

Explanation:

6 0
2 years ago
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
babunello [35]
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.


Read more on Brainly.com - brainly.com/question/18743384#readmore
8 0
3 years ago
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