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weeeeeb
1 month ago
8

The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

friction between each block and the surface is 0.4 . The two blocks collide, stick together, and move to the right. Remember that the spring is not attached to the 8 kg block. Find the speed of the 8 kg block just before it collides with the 2 kg block. Answer in units of m/s.
Physics
1 answer:
kicyunya [3.2K]1 month ago
7 0
3.258 m/s Explanation: The spring constant is assumed to be 263 N/m and the displacement of the spring is also assumed to be 0.7 m; the coefficient of friction between blocks is 0.4. The energy stored in the spring is described by . Given the conservation of energy in the system, the speed of the 8 kg block just prior to collision is 3.258 m/s.
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Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is \sqrt{725} ft/ sec

Clarification:

Provided:

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h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Now, let's calculate the derivative of distance in relation to time

\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

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A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/
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A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a
Softa [3030]

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Explanation:

Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

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Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

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The car moves at a constant speed in this direction, indicating that its acceleration is zero.

a=0

Thus, according to Newton's second law,

F=ma

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F=0

Learn more about slopes and friction:

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