The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.
<h3>Work done by the applied force</h3>
The area under force versus displacement graph is work done.
The total work done by pulling the object 7 m, can be grouped into two areas;
- First area, A1 = area of triangle from 0 m to 2.0 m
- Second area, A2 = area of trapezium, from 2.0 m to 7.0 m
A1 = ¹/₂ bh
A1 = ¹/₂ x (2) x (20)
A1 = 20 J
A2 = ¹/₂(large base + small base) x height
A2 = ¹/₂[(7 - 2) + (7-3)] x 50
A2 = ¹/₂(5 + 4) x 50
A2 = 225 J
<h3>Total work done </h3>
W = A1 + A2
W = 20 J + 225 J
W = 245 J
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The total resistance is 420 ohm.
A circuit with resistive elements of 220, 100, 57, and 43 produce what total resistance
R= 220+ 100+ 57+ 43
= 420 Ω
What is resistance and its types?
Resistance is a measure of the opposition to current flow in an electrical circuit also known as ohmic resistance or electrical resistance. Ohms are measured as resistance, symbolized by the Greek letter omega (Ω). The ratio of the applied voltage to the current through the material is then known as resistance.
What causes resistance?
An electric current flows when electrons move through a conductor, such as a metal wire. The moving electrons can collide with the ions in the metal. This makes it more difficult for the current to flow, and causes resistance.
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Answer:
Net forces which pushes the window is 30342.78 N.
Explanation:
Given:
Dimension of the office window.
Length of the window = 
m
Width of the window = 
m
Area of the window = 
Difference in air pressure = Inside pressure - Outside pressure
= 
atm = 
atm
Conversion of the pressure in its SI unit.
⇒ 
atm = 
Pa
⇒ atm = 
Pa
We have to find the net force.
We know,
⇒ Pressure = Force/Area
⇒ 
⇒ 
⇒ Plugging the values.
⇒ 
⇒ 
Newton (N)
So,
The net forces which pushes the window is 30342.78 N.
Answer: a) 139.4 μV; b) 129.6 μV
Explanation: In order to solve this problem we have to use the Ohm law given by:
V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.
Then we have:
for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω
and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω
Finalle we calculate the potential difference (V) for both wires:
Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V
V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V
