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Here is the full information about the question. <span>Ilya and Anya each can run at a speed of 8.50mph and walk at a speed of 3.50 mph . They set off together on a route of length 5.00 miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
the calculation would be:
</span><span>
t = d / s </span>
<span>t = 2.5 (half of the total distance) / 8.5 (speed of running) </span>
<span>This is .294 hours which is about 1058s... </span>
<span>for the walking part... </span>
<span>t = d / s </span>
<span>t = 2.5 / 3.5 </span>
<span>t = 5/7hours = 2571 s. </span>
Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal
Answer:
4500 N
Explanation:
When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:
a = v^2/r
where v is the velocity of the body and r is the radius of the circumference:
Therefore, a body with mass m, will feel a force f:
f = m v^2/r
Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:
fs = μN = μmg
The car will not slide if f = fs, i.e.
fs = μmg = m v^2/r
That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration
fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N
