The friction force exerted on the 2400kg car is 9408 Newtons
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What force will be exerted on the 2400kg car?</h3>
First, we know that the friction force and mass are represented by a proportional relation, this means that we can write:
F = k*M
Where F is the force, M is the mass and k is the constant of proportionality.
We know that for a 1600kg car, a force of 6272N is exerted, replacing that we get:
6272N = k*1600kg
Solving for k we get:
k = (6272N)/(1600 kg) = 3.92 N/kg
Then the proportional relationship is:
F = (3.92 N/kg)*M
So if M = 2400kg, we have:
F = (3.92 N/kg)*2400kg = 9408 N
So the friction force exerted on the 2400kg car is 9408 Newtons
If you want to learn more about proportional realtions:
brainly.com/question/12242745
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Find slope: (y2-y1)/(x2-x1)
(6-3)/(-5-0) = 3/-5 = -3/5
Equation: y = -3/5x + b
Plug in a point
3 = -3/5(0) + b, b = 3
Solution: y = -3/5x + 3
We have two widths of the same length plus one length
Let the width be
and the length be
Perimeter = 2×width + 2×length
(equation 1)
The cost is $5 per foot on the width and $25 per foot on the length
Total cost = (5 × 2 × width) + (25 × length)
(equation 2)
We have two variables that we need to solve, so we will need to use the simultaneous equations method (either elimination or substitution)
Since equation 1 is given
, we can rearrange the equation to make
the subject
then substitute this into equation 2
Substitute w = 75 back into 190 = w + l
190 = 75 + l
l = 190 - 75
l = 115
Answer:
Length = 115 feet
Width = 75 feet