A. Impulse is simply the product of Force and time.
Therefore,
I = F * t --->
1
where I is impulse, F is force, t is time
However another formula for solving impulse is:
I = m vf – m vi --->
2
where m is mass, vf is final velocity and vi is initial
velocity
Therefore using equation 2 to solve for impulse I:
I = 2000kg (0) – 2000kg (77 m/s)
I = -154,000 kg m/s
B. By conservation of momentum, we also know that Impulse
is conserved. That means that increasing the time by a factor of 3 would still
result in an impuse of -154,000 kg m/s. So,
I = F’ * (3 t) = -154,000 kg m/s
Since t is multiplied by 3, therefore this only means
that Force is decreased by a factor of 3 to keep the impulse constant,
therefore:
(F/3) (3t) = -154,000 kg m/s
Summary of Answers:
A. I = -154,000 kg m/s
B. Force is decreased by factor of 3
Answer:
ΔK.E = 14 nJ
Explanation:
Solution:
- The charge that moves under the influence of an Electric Field produced between a potential difference (V) stores electric potential energy U within that is converted to kinetic energy.
- We will use conservation of energy on the system that contains the charged particle with charge q loses its electric potential energy U as it moves towards positively charged object that converts into a gain in Kinetic energy of the charged particle ΔK.E:
ΔK.E = U
Where,
U = V*q
ΔK.E = V*q
ΔK.E = (7*10^-6)*(2*10^-3)
ΔK.E = 14 nJ
- The gain in kinetic energy is 14 nJ.
6 and 
Explanation:
The expression:
6a - 
A coefficient is a numerical constant usually placed before a variable in an expression.
They are used to multiply variables.
Given equation is;
There are two variables in this expression which are;
a and - b
The coefficients in this expression are:
6 and
learn more:
Quadratic equation brainly.com/question/1357167
#learnwithBrainly
Answer:
Because in order for work to be done on an object, the object must be moving. Why isn't work being done on a barbell when a weight lifter is holding the barbell over his head? Work is maximized when force is applied in the same direction that the object is moving. ... In order to do work faster, more_is required.
It's gravitational potential energy at the top will roughly equal it's kinetic energy when it was released (a little is lost to air resistance). Note this will assume the release point is zero potential energy. (we are free to define it that way, just letting you know). Gravitational potential energy is mgh.
mgh=25J
h=25J/(0.5kg x 9.81m/s^2) = 5.097m
So it goes about 5.1 meters above the point where it was released
