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velikii [3]
3 years ago
14

Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity v0 and glider

B is initially at rest. After they collide, A has a velocity 12v0 and B has a velocity 92v0. Find the ratio of the gliders' masses, mAmB.
Physics
1 answer:
LenaWriter [7]3 years ago
3 0

Answer:

Explanation:

Applying conservation of momentum

mA v0 + 0 = mA x 12v0 + mB x 92v0

11 mA = - 92mB .

mA / mB = 92 / 11

mA : mB = 92 : 11.

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10 turns of wire are closely wound around a pencil as shown in the figure. when measured using a scale as shown, the length of t
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Answer:

a. The thickness of the wire is 2.5 mm.

b. The wire is 0.25 cm thick.

Explanation:

Number of turns of the wire = 10

The length of total turns = 25 mm

a. The thickness of the wire can be determined by;

thickness of the wire = \frac{length of total turns}{number of turns}

                           = \frac{25}{10}

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Therefore, the wire is 2.5 mm thick.

b. To determine the thickness of the wire in centimetre;

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So that,

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8 0
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Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not
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Answer:

Part A: t = v_0/a_0

Part B: t = v_0/a_0

Part C: v_0^2/a_0

Explanation:

Part A:

We will use the following kinematics equation:

v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}

Part B:

We will use the same kinematics equation:

v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}

Part C:

The total time takes is 2t.

So the train moves a distance of

x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}

And the car moves a distance in Part A and in Part B:

d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}

So the total distance that the car traveled is d = \frac{v_0^2}{a_0}

The difference between the train and the car is

x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}

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