Physics, 20.09.2019 03:20 mem8163

Ared car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis.,theredcarisatxr 220 m. if the red car has a constant velocity of 20 km/h, the cars pass each other at x 44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x 76.6 m. what are (a) the initial velocity and (b) the constant acceleration of the green car?

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Answer from: rcmolyneaux

a) The initial velocity of the green car is -13 m/s

b) The acceleration of the green car is - 2.25 m/s²

Explanation:

The equation for the position of objects moving in a straight line with constant acceleration is as follows:

x = x0 + v0·t + 1/2·a·t²

where:

x = position

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

If the velocity is constant, then a = 0 and x = x0 + v·t

a)The initial position of the red and green car is 0 m and 220 m respectively. We know that at 44.5 m the cars pass each other if the red car has a constant velocity of 20 km/h. So let´s find how much time it takes the cars to pass each other in this case:

The position of the red car is:

x = x0 + v·t

then:

0.0445 km = 0 km + 20 km/h · t

t = 0.0445 km/ 20 km/h = 8.0 s

We also know that if the red car has a velocity of 40 km/h, both cars pass each other at 76.6 m. So let´s find the time it takes the cars to reach that position using the equation for the red car:

0.0766 km = 0 km + 40 km/h · t

t = 0.0766 km / 40 km/h = 6.9 s

The position of the green car at t= 6.9 s and t = 8.0 s must be the same as the red car because both cars pass each other at those times.

Then, for the green car:

x = x0 + v0·t + 1/2·a·t²

0.0445 km = 0.220 km + v0 · 8.0 s + 1/2·a· (8.0 s)²

and

0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·a· (6.9 s)²

Now we have a system of two equations with two unknowns.

Solving for "a" in the first equation

0.0445 km - 0.220 km - v0 · 8.0 s = 32 s²·a

(-0.176 km - v0 · 8.0 s) / 32 s² = a

Replacing a = (-0.176 km - v0 · 8.0 s) / 32 s² in the second equation and solving for v0:

0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·((-0.176 km - v0 · 8.0 s)/32 s²)·(6.9 s)²

-0.143 km = v0 · 6.9 s - 0.74(0.176 km + v0 · 8.0 s)

-0.143 km = v0 · 6.9 s - 0.130 km - v0 · 5.9 s

-0.143 km + 0.130 km = v0 · 6.9 s - v0 · 5.9 s

-0.013 km = 1 s · v0

v0 = -13 m/s

b) The acceleration of the green car is:

a = (-0.176 km - v0 · 8.0 s) / 32 s²

a = (-0.176 km - (-0.013 km/s) · 8.0 s) / 32 s² = -2.25 m/s²

Answer from: Quest

heats surface water up

Answer from: Quest

The answer is 4f======since electrostatic force is given by $f_{elec} = \dfrac{kq_1 q_2}{r^2}$ we let f be the electrostatic force when r = 1 m $f = \dfrac{kq_1 q_2}{1^2} = kq_1q_2$ if we let $f_{0.5}$ denote the electrostatic force when the distance is decreased so that r = 0.5m, then $\begin{aligned} f_{0.5} & = \dfrac{kq_1 q_2}{0.5^2} \\ & = \frac{1}{0.5^2} \cdot kq_1q_2 \\ & = \frac{1}{\left( \frac{1}{2}\right)^2} \cdot kq_1q_2 \\ & = 4 \cdot \boxed{ kq_1q_2 } \end{aligned}$ we established earlier that f = kq₁q₂. therefore $f_{0.5} = 4 f$ the force will now be 4f.

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