Answer:
A.) the inverse of the square of the distance separating them
Explanation:
Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."
Mathematically, F = kq1q2/r²
Where q1 and q2 are the charges
r is the distance between the charges.
According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.
<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’):
</em></h2><h2><em>
</em></h2><h2><em>The car’s speed is 25m/s
</em></h2><h2><em>The distance travelled is 75m
</em></h2><h2><em>Then we have the formulas for speed and distance:
</em></h2><h2><em>
</em></h2><h2><em>v = a x t -> 25 = a x t
</em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2
</em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say:
</em></h2><h2><em>
</em></h2><h2><em>t = 25 / a
</em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a
</em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1):
</em></h2><h2><em>
</em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2
</em></h2><h2><em>t^2 = 150 / a
</em></h2><h2><em>Since t has the same value for both truths we can say:
</em></h2><h2><em>
</em></h2><h2><em>625 / a^2 = 150 / a
</em></h2><h2><em>
</em></h2><h2><em>Thus multiply both sides with a^2:
</em></h2><h2><em>
</em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17
</em></h2><h2><em>
</em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW
Answer:
A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force
Answer:
mass goes down volume remains the same
