Answer:
increase.
Explanation:
According to the newton’s second law of motion force is expressed as product of mass and acceleration.
F = m a
If the force acting is constant, then.
m∝ 
That is if the mass of object increases the acceleration decreases and vice versa. The above equation is used when the force acting on the body is constant.
As the thrust force from the rocket engine is constant throughout there will be a variation in the mass or acceleration.
Thus, it won't stay the same.
As the weight of the car is maximum at the start because of the fuel present in the rocket engine and minimum at the end as the fuel burns throughout the journey of the car. Weight will be minimum at the end and hence acceleration is maximum at the end.
Thus, it won't decrease.
As the acceleration is going from minimum at the start to maximum at the end, therefore it is continuously increases throughout its journey.
Thus, it will increase.
I would say the answer is 3 because by falling technically the ball would be kind of moving in the air. Plus potential energy is when for example a soccer ball isnt moving
The change in potential energy of this system is = 40 J.
<h3>What is the potential difference?</h3>
The potential difference, often known as voltage, is equal to the amount of current times the resistance. One Joule, or one Volt, of energy is required for one Coulomb of charge to flow from one place in a circuit to another.
<h3>What is the formula for potential difference?</h3>
V=Uq The change in potential energy of a charge q transported from point A to point B, divided by the charge, is what is used to determine the electric potential difference between points A and B, or VBVA. The joules per coulomb unit of potential difference is called the volt (V).
This system's potential energy changed by 40 J.
Energy = charge * potential difference
=> Energy = (8v)*5
=> Energy = 40 J
The change in potential energy of this system is = 40 J.
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the electrons should be in the outer valence levels/shells.
Answer:
x(t)=0.337sin((5.929t)
Explanation:
A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation
Definition of parameters
m=mass 3kg
k=force constant
e=extension ,m
ω =angular frequency
k=90/1.6=56.25N/m
ω^2=k/m= 56.25/1.6
ω^2=35.15625
ω=5.929
General solution will be
differentiating x(t)
dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)
when x(0)=0, gives c1=0
dx(t0)=2m/s gives c2=0.337
Therefore, the position of the mass after t seconds is
x(t)=0.337sin((5.929t)
