<h3><u>Answer</u>;</h3>
-The total momentum of an isolated system is constant.
-The total momentum of any number of particles is equal to the vector sum of the momenta of the individual particles.
-The vector sum of forces acting on a particle equals the rate of change of momentum of the particle with respect to time.
<h3><u>Explanation</u>;</h3>
- Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momenta of the multiple bodies which make up a system.
- The vector sum of forces acting on a particle is equivalent to the rate of change of momentum of the particle with respect to time. This is according to the Newton's second Law of motion. In mathematical terms, ֿF = d ֿp/dt, that is F= ma.
- According to the Law of conservation of Momentum, or a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
Option A, current (thumb) to magnetic field (fingers)
As per the First right-hand rule,
Using right hand, if we suppose that thumb points towards the electric current
fingers curl towards the magnetic field
I mean if he flies 5g that means that's his average speed too
To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
Here,
v = Final velocity
= Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
Now the position can be calculated as,
When it has the ground, y=0 and the time is t=4s,
Therefore the cliff was initially to 41.6m from the ground
Answer:
P₁- P₂ = 91.1 10³ Pa
Explanation:
For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)
In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference
For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m
P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)
P₁-P₂ = 22.5 10³ + 68.6 10³
P₁- P₂ = 91.1 10³ Pa
