The change in momentum is 5500 kg m/s
Explanation:
The change in momentum of an object is given by

where
m is the mass of the object
v is the final velocity
u is the initial velocity
In this problem, we have:
(mass of the motorcycle)
(final velocity)
(initial velocity)
Therefore, the change in momentum is

Learn more about change in momentum:
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Answer:
- apples falling from trees
- people's feet touching the ground
- sky divers moving toward the ground
- balls bending downward after being thrown
Explanation:
When a space ship is accelerating in space, there is a force known as Inertia that kicks in. Inertia will mirror the effects of gravity on the ship even if there is no gravitational field effect such that anything that would happen where there is gravity, would continue to happen.
This means that apples will fall from trees, people's feet will touch the ground, sky divers will be pulled downwards and balls will bend downwards when thrown as well. These are the same effects expected on earth where gravity pulls things towards the earth's core.
Answer:
so, why we have to find here..
<h3>stay safe healthy and happy.</h3>
Using Newton's second law of motion:
F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)
Given: Find: Formula: Solve for m:
F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2
a= 200m/s^2
Solution:
Using Eq.2
m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg
Answer:

Explanation:
<u>LC Circuit</u>
It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:
= charge of the capacitor in any time 
= initial charge of the capacitor
=angular frequency of the circuit
= current through the circuit in any time 
The charge in an LC circuit is given by

The current is the derivative of the charge

We are given

It means that
![q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]](https://tex.z-dn.net/?f=q%28t_1%29%20%3D%20q_0%20%5C%2C%20cos%20%28%5Comega%20t_1%20%29%3Dq_1%5C%20.......%5Beq%201%5D)
![i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]](https://tex.z-dn.net/?f=i%28t_1%29%20%3D%20-%20%5Comega%20q_0%20%5C%2C%20sin%28%5Comega%20t_1%29%3Di_1.........%5Beq%202%5D)
From eq 1:

From eq 2:

Squaring and adding the last two equations, and knowing that


Operating

Solving for 

Now we know the value of
, we repeat the procedure of eq 1 and eq 2, but now at the second time
, and solve for 

Solving for 

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.




Finally

