D. Advances in technology were needed to gather more evidence.
Explanation:
New technologies had to be developed to ascertain Wegener's claims.
The major flaw in Wegener's postulate was that the moving continents lacked a mechanism to drive them into motion.
His theory needed more scientific backing also.
- A major leap was made when new technologies used during the second world war was deployed to investigate the ocean floor.
- Sequences of magnetic anomalies were unraveled using equipment developed during the second world war
- This gave further proof to the idea of continental drift and it expanded the theory into a more broader and inclusive theory of plate tectonics.
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amount of work done is 5880 J
Given:
mass of object = 50kg
Final height = 20m
initial height = 8m
To Find:
amount of work done
Solution:
work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.
The work done by gravity is given by the formula,
W = mgh
W = 50 x 9.8 x ( 20-8)
= 5880 J
So the work done is 5880 J
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The energy is 3.06 electronvolts, E = 3.06eV
1eV = 1.6 * 10^-19 J
3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J
Answer:
The mass of the man is 71 kg
Explanation:
Given;
kinetic energy of the man, K.E = 887.5 J
velocity of the man, v = 5 m/s
The mass of the man is calculated as follows;
K.E = ¹/₂mv²
where;
m is the mass of the man
2K.E = mv²
m = 2K.E / v²
m = (2 x 887.5) / (5)²
m = 71 kg
Therefore, the mass of the man is 71 kg
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
