6 days ago

A 0.75 kg wood block is pushed 0.2 m into a horizontal spring with a spring constant of 50 N/m and released. a. How much spring

potential energy is in the system before the wood block begins to move? b. If the surface is frictionless, what is the kinetic energy of the block as it leaves the spring? c. If the surface is frictionless, what is the velocity of the block as it leaves the spring? d. Assuming the surface is not frictionless, and the wood block leaves the spring at a velocity of 1.32 m/s. How much energy is lost due to friction?

Physics

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Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads north, at 8 AM, and walks steadily at 2 mi

Keith_Richards [3271]

Answer:

2.57 hours

Explanation:

Let t (in hours) represent the time it takes for Elsie to walk until they are separated by 25 miles. Since Steve starts 2 hours earlier, his time will be t + 2.

The distance covered by Steve heading north is $s_s = 2(t + 2)$

The distance Elsie travels heading west is $s_e = 2.5t$

The separation between Steve and Elsie is

$\sqrt{s_s^2 + s_e^2} = \sqrt{(2(t+2))^2 + (2.5t)^2} = 25$

We can solve for t by squaring both sides.

$(2(t+2))^2 + (2.5t)^2 = 25^2 = 625$

$4(t+2)^2 + 6.25t^2 = 625$

$4(t^2 + 4t + 4) + 6.25t^2 = 625$

$10.25t^2 + 16t - 609 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{-16\pm \sqrt{(16)^2 - 4*(10.25)*(-109)}}{2*(10.25)}$

$t= \frac{-16\pm68.74}{20.5}$

Therefore, t = 2.57 or t = -4.13.

Since t cannot be negative, we select t = 2.57 hours.

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A stagehand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N. W

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The answer is option (a). A stagehand applies a horizontal force of 176 N to start moving a large piece of scenery that was initially at rest. The equation to determine this force has to do with its rotational speed and centripetal acceleration.

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Torque can be calculated by multiplying the force (N) applied at 90∘ to the lever arm at a distance (m) from the pivot point (po

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Torque = Force * Distance perpendicular

Torque = 15 N * 2 m = 30 Nm

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