Answer: 0.5 m
Explanation:
Given
Mass of the person is 
Trampoline launches the person into the air up to height of 
Force experience by springs is 
Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.
![\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m](https://tex.z-dn.net/?f=%5CRightarrow%20F%5Ccdot%20x%3Dmgh%5Cquad%20%5B%5Ctext%7Bx%3Ddisplacement%20of%20the%20trampoline%7D%5D%5C%5C%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B50%5Ctimes%209.8%5Ctimes%202%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B980%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D0.5%5C%20m)
The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2) where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.
Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span>
From reliable sources in the internet, the half-live of carbon-14 is given to be 5,730 years. In a span of 10,000 to 12,000 years, there are almost or little more than 2 half-lives. Thus, there should be
A(t) = A(0)(1/2)^t
where t is the number of half-lives, in this case 2. Thus, only about 1/4 of the original amount will be left.
Melting, as igneous rock is magma or lava that has cooled and hardened.
Answer:
20,850 N
Explanation:
We can solve the problem by using second Newton's Law:
where
F is the force
m is the mass
a is the acceleration
In this problem, we have:
m = 70 kg is the mass
is the acceleration (which is negative, because it is a deceleration)
So, we can use the equation above to find the force:
and the negative sign simply means that the force is in the opposite direction to the motion.
