Answer:
Hello there use something that looks like this
Explanation:
This is an accurate representation of something you are working on!
As you can see the wire and the core are represented on the left and is showing how it can be represented on your right hand and how they are similar!
Answer:
The acceleration is 2.448 meters per square second and is vertically upward.
Explanation:
The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:
(1)
Where:
- Buoyant force, measured in newtons.
- Mass of the plastic ball, measured in kilograms.
- Gravitational acceleration, measured in meters per square second.
- Net acceleration, measured in meters per square second.
If we know that ,
and
, then the net acceleration of the plastic ball is:
The acceleration is 2.448 meters per square second and is vertically upward.
Answer:
the resistance of the wire has no effect on the brightness of the bulb.
Explanation:
Let's apply ohm's law for your light bulb circuit plus wires plus switch
V = I R_{bulb} + I R_ {wire}
the current in a series circuit is constant
V = I (R_{bulb} + R_{wire})
To know the effect of the wires on the brightness of the bulb, we must look for the value of the typical resistance of these elements.
Incandescent bulb
Power 60 W
let's use the power ratio
P = V I = V2 / R
R = V2 / P
the voltage value for this power is V = 120 V
R = 120 2/60
R_bulb = 240 Ω
Resistance of a 14 gauge copper wire (most used), we look for it on the internet
R = 8.45 Ω/ km
in a laboratory circuit approximately 2 m is used, so the resistance of our cable is
R = 8.45 10⁻³ 2
R_wire = 0.0169 Ω
let's buy the two resistors
R_{bulb} = 240
R_{wire} = 0.0169
therefore resistance of the bulb is much greater than that of the wire, therefore almost all the power is dissipated in the bulb.
In summary, the resistance of the wire has no effect on the brightness of the bulb.