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monitta
3 years ago
5

A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the

same point and accelerates uniformly at 4.0 m/s2 toward the north. How long after the second car starts does it overtake the first car? O 12 s O 19 s O 21 s o 24 s
Physics
1 answer:
yKpoI14uk [10]3 years ago
6 0

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

       x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

        x = ½ a₂ (t-4)²

Let's solve

       a₁  t² = a₂ (t-4)²

      a₁/a₂ t² = t² -2 4 t + 16

      t² (1- 2.0 / 4.0) - 8 t +16

      t² 0.5 - 8 t +16 = 0

      t² -16 t + 32 = 0

Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

     t = ½ (16 ± 11,3)

Solutions

     t1 = 13.66 s

     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

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You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

I_i = Initial moment of inertia = 9.3 kgm²

I_f = Final moment of inertia = 5.1 kgm²

\omega_i = Initial angular speed = 1.8 rev/s

\omega_f = Final angular speed

As the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

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Which of the following is a good conductor of heat? 1. water 2. glass 3. wood 4. mercury​
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Explanation:

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Read 2 more answers
Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters
amid [387]

Answer:

Part a)

t = 1.65 s

Part b)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

v = 27.3 m/s

direction of velocity is given as

[tex]\theta = 26.35 degree

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

a = \frac{3}{4}(9.81)

a = 7.36 m/s^2

now the vertical displacement covered by the canister is given as

y = 10 m

now by kinematics we have

y = \frac{1}{2}gt^2

10 = \frac{1}{2}(7.36)t^2

t = 1.65 s

Part b)

Horizontal speed of the canister is given as

v_x = 24.5 m/s

now the distance moved by it

x = v_x t

x = 24.5 (1.65)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

v_x = 24.5 m/s

final velocity in Y direction

v_y = v_i + at

v_y = 0 + (7.36)(1.65)

v_y = 12.14 m/s

now magnitude of velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{24.5^2 + 12.14^2}

v = 27.3 m/s

direction of velocity is given as

\theta = tan^{-1}\frac{v_y}{v_x}

\theta = tan^{-1}\frac{12.14}{24.5}

[tex]\theta = 26.35 degree

6 0
3 years ago
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