Complete question:
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.
Answer:
The electric field inside this metal resistor is 3125 V/m
Explanation:
Given;
length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m
diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m
the potential difference due to electric field between the two ends of the resistor, V = 10 V
The electric field inside this metal resistor is given by;
ΔV = EL
where;
ΔV is change in electric potential
E = ΔV / L
E = 10 / (3.2 x 10⁻³ )
E = 3125 V/m
Therefore, the electric field inside this metal resistor is 3125 V/m
When developing an experimental design, the action that would improve the quality of the results is to ensure that it answers a question about cause and effect.
<h3>What is experimental design?</h3>
Experimental design is a concept used to organize, conduct, and interpret results of experiments in an efficient way, making sure that as much useful information as possible is obtained by performing a small number of trials.
Thus, when developing an experimental design, the action that would improve the quality of the results is to ensure that it answers a question about cause and effect.
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Explanation:
Why are high tides found simultaneously on opposite sides of Earth? The tidal bulges occur on both sides of Earth that are aligned with the tide-generating body. The ocean water experiencing high tide rotates around Earth on a 12-hour cycle.
The coefficient of friction between the road and the car's tire is determined as 0.78.
<h3>Acceleration of the car</h3>
The acceleration of the car is calculated as follows;
v² = u² - 2as
0 = u² - 2as
a = u²/2s
where;
- u is the initial velocity = 97 km/h = 26.94 m/s
a = (26.94)²/(2 x 47)
a = 7.72 m/s²
<h3>Coefficient of friction</h3>
μ = a/g
μ = (7.72)/9.8
μ = 0.78
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In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>