Answer:— A square labeled Object B is shown with an arrow pointing up labeled 3 N, an arrow pointing right labeled 2 N, an arrow pointing down labeled 3 N and an arrow pointing left labeled 1 N. In which direction will the objects move? Object A will move down and Object B will move up.
Explanation:
Answer:

Explanation:
We'll use the momentum equation:

where:
p = momentum
m = mass
v = velocity
Since we're doing the magnitude of momentum of the system, we'll add the mass of the cyclist and the mountain bike together:

Given that, we can now substitute our given values into the momentum equation:

Our final answer is:

For this one, all you really need to do is eliminate any answers
that are absurd or meaningless.
You can't increase a transformer.
You can't increase a circuit.
You can't increase a generator.
When the <em><u>current</u></em> through a coil of wire increases, the magnetic field
around the coil increases, so there would be more magnetic force
between the coil and a permanent magnet.
Increasing his acceleration will impact his velocity and rate of displacement covered in that as the speed increases due to the increased rate of acceleration, the rate of air resistance also increases.
<h3>What is air resistance?</h3>
Air resistance is a force created by air. When an item moves through the air, the force operates in the opposite direction.
When a diver descends, the force of air resistance acts to counteract the force of gravity. As the skydiver falls faster and faster, the quantity of air resistance grows until it equals the magnitude of gravity's force.
A balance of forces is achieved when the force of gravity equals the force of air resistance, and the skydiver no longer accelerates. The skydiver reaches what is known as terminal velocity.
Learn more about air resistance:
brainly.com/question/16859536
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Answer: I = 111.69 pA
Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.
The hall voltage of a semiconductor sensor is given below as
V = I×B/qnd
Where V = hall voltage = 1.5mV =1.5/1000=0.0015V
I = current =?,
n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3
q = magnitude of an electronic charge=1.609×10^-19c
B = strength of magnetic field = 5T
d = thickness of sensor = 0.8mm = 0.0008m
By slotting in the parameters, we have that
0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008
0.0015 = I×5/7.446×10^-8
I = (0.0015 × 7.446×10^-8)/5
I = 111.69*10^(-12)
I = 111.69 pA