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jasenka [17]
3 years ago
13

A thief is trying to escape from a parking garage after completing a robbery, and the thief's car is speeding (v = 13 m/s) towar

d the door of the parking garage (fig. p2.60). when the thief is l = 22 m from the door a police officer flips a switch to close the garage door. the door starts at a height of 9 m and moves downward at 0.6 m/s. if the thief's car is 1.4 m tall, will the thief escape?
Physics
1 answer:
rodikova [14]3 years ago
3 0

<u>Answer</u>

Yes, the car reaches the door before the gate closes.

<u>Explanation</u>

The time taken by the car to reach at the door.

Time = distance / time

        = 22/13

          = 1.6923 seconds

Time taken by the door to close up to the height of the car.

Distance the door has to move to prevent the car from escaping = 9.1.4 = 7.6 m

From newton's 2nd law of motion;

s = ut + 1/2 gt²

7.6 = 0.6t + 1/2 × 10t²

7.6 = 0.6t + 5t²

50t² + 6t - 76 = 0

Solving this quadrilatic equation,

t = 28.537 seconds

Answer: Yes, the car reaches the door before the gate closes.


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MAVERICK [17]

Answer:

 E = 1,873 10³ N / C

Explanation:

For this exercise we can use Gauss's law

        Ф = E. dA = q_{int} / ε₀

Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.

The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.

        The surface of a sphere is

             A = 4π r²

             E 4π r² = q_{int} /ε₀

  The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is

           q_{int} = q₁ + q₂

           q_{int} = (530 - 200) 10⁻⁹

           q_{int} = 330 10⁻⁹ C

The electric field is

             E = 1 / 4πε₀   q_{int} / r²

            k = 1 / 4πε₀

            E = k q_{int}/ r²

Let's calculate

           E = 8.99 10⁹   330 10⁻⁹/ 1.32²

           E = 1,873 10³ N / C

7 0
3 years ago
An object that is speeding up is accelerating.<br> True<br> False
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True. The speed of any object will have faster acceleration, and eventually slow down due to gravity.
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Match each chemical name with the correct chemical formula.
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8 0
3 years ago
(ASAP) would it be 125 m/s2 to calculate for her speeding up?
serg [7]

Answer:

0\:\mathrm{ m/s^2}

Explanation:

Recall the formula for acceleration:

\displaystyle\\a=\frac{v_f-v_i}{\Delta t}, where v_f is final velocity, v_i is initial velocity, and \Delta t is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0
2 years ago
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