Answer:
ΔF=125.22 %
Explanation:
We know that drag force on the car given as
=Drag coefficient
A=Projected area
v=Velocity
ρ=Density
All other quantity are constant so we can say that drag force and velocity can be given as
Now by putting the values
Percentage Change in the drag force
ΔF=125.22 %
Therefore force will increase by 125.22 %.
<span>computing or networking is a distributed application architecture that partitions tasks or work loads between peers. Peers are equally privileged, equipotent participants in the application. They are said to form a peer-to-peer network of nodes.</span>
Based on the options given, the most likely answer to this query is C) 4577 liters.
Upon computation of the given variables the result seems to be 4577 L
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Answer:
Explanation:
Using the equation of motion v² = u²+2as
v is the final velocity = 40m/s
u is the iniyail velocity 0m/s
a is the acceleration
s is the displacement
Substituting in the formula;
40² = 0²+2a(50)
1600 = 100a
Divide both sides by 100
100a/100 = 1600/100
a = 16
Hence the car acceleration is 16m/s²
R = 0.407Ω.
The resistance R of a particular conductor is related to the resistivity ρ of the material by the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of the material.
To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.
We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4. Then:
R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]
R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²
R = 0.407Ω
