Answer:
He could jump 2.6 meters high. 
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense. 
 
        
             
        
        
        
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s
 
        
             
        
        
        
A convergent meniscus lens is a lens that is composed of two spherical surfaces, like the on shown next:
The imaginary line that runs through the middle of the lens is the "symmetry axis". 
In this type of lenses incident parallel beams of light converge in one point, as follows:
And thus we get the diagram. 
 
        
             
        
        
        
By using the second law of Newton, the frictional force is 200N.
We need to know about the second law of Newton (force) to solve this problem. The total force applied an object is proportional to the mass of object and acceleration. It can be defined as
∑F = m . a
where F is force, m is mass and a is acceleration.
From the question above, we know that
F1 = 200N
v = constant therefore (a = 0 m/s²)
By using second law of Newton, we get
∑F = m . a
F1 - Ffriction = m . 0
200 - Ffriction = 0
Ffriction = 200 N
Hence, the frictional force is 200N. 
Find more on force at: brainly.com/question/25239010
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