Answer:
<em>The object-Earth system is open</em>

Explanation:
<u>Accelerated Motion
</u>
When an object is released in free air (with no other forces than the gravity), it describes a free-fall motion and the formulas include the acceleration of gravity as part of the calculations. But when there is another external force, then the acceleration is not the gravity, but the result of the net force exerted on the mass of the object.
By definition, an open system includes the exchange of energy from and to the surroundings, that is why all systems surrounding our planet are considered as open systems. In our case, the object is interacting with the planet's gravity and there is some other external force, which will be computed later. The object-Earth system is open.
If the object starts from rest, its initial speed is zero, and

where a is the acceleration and t is the time. The distance traveled is given by
:

From the two above equations, we find that:

Solving for a



It means the net force is

The object's weight is

This means there is some external force acting upwards delaying the object's fall of a magnitude of

Answer:
a)equal to
Explanation:
properties of a wave
applies in reflection
others include:refraction,diffraction,interference
The series circuit has the higher equivalent resistance, but to find the larger equivalent resistance you have to use Ohm's law (V2=I2R2), the power dissipated by the resistor can also be found using P2=I22R2=V22R2. To find the equivalent resistance of the circuit, notice that the parallel connection of R2 and R3 is in series with R1, so the equivalent resistance is Req=R1+(1R2+1R3)−1=1.00Ω+(16.00Ω+113.00Ω)−1=5.10Ω.
Hope this helps!
Please give Brainliest!
Answer:
The current in the second loop will stay constant
Explanation:
Since the induced emf in the second coil, ε due to the changing current i₁ in the first wire loop ε = -Mdi₁/dt where M = mutual inductance of the coils and di₁/dt = rate of change of current in the first coil = + 1 A/s (positive since it is clockwise)
Now ε = i₂R where i₂ = current in second wire loop and R = resistance of second wire loop.
So, i₂R = -Mdi₁/dt
i₂ = -Mdi₁/dt/R
Since di₁/dt = + 1 A/s,
i₂ = -Mdi₁/dt/R
i₂ = -M × + 1 A/s/R
i₂ = -M/R
Since M and R are constant, this implies that i₂ = constant
<u>So, the current in the second wire loop will stay constant.</u>