Answer: Alfred Wegener provided some of the important points that supported the theory of continental drift. They are as follows-
- The continents were once all attached together, and this can be proved by studying the coastlines of some of the continents that perfectly matches with one another.
- The appearance of similar rock types and similar fossils (including both animals and plants) has also contributed much information that continents were once all together.
Answer:
this is what popped up when I searched it up:In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.
Explanation:
X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2
![T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2usin%20%5Ctheta%7D%7B%5Cfrac%7BEq%7D%7Bm%7D%7D%20%3D%5Cfrac%7Bm%2A%202usin%20%5Ctheta%7D%7BEq%7D)
Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;
![T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B%281.67X10%5E%7B-27%7D%2A%202%2A3X10%5E4sin%2030%7D%7B400%2A1.6X10%5E%7B-19%7D%7D)
T = 7.83 X10⁻⁷ s
Answer:
The efficiency is 0.33, or 33%.
Explanation:
From the thermodynamics equations, we know that the formula for the efficiency of a heat engine is:
![\eta=1-\frac{Q_2}{Q_1}](https://tex.z-dn.net/?f=%5Ceta%3D1-%5Cfrac%7BQ_2%7D%7BQ_1%7D)
Where η is the efficiency of the engine, Q_1 is the heat energy taken from the hot source and Q_2 is the heat energy given to the cold object. So, plugging the given values in the formula, we obtain:
![\eta=1-\frac{200J}{300J}=0.33](https://tex.z-dn.net/?f=%5Ceta%3D1-%5Cfrac%7B200J%7D%7B300J%7D%3D0.33)
This means that the efficiency of the heat engine is 0.33, or 33% (The efficiency of an engine is dimensionless).