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Nataly_w
1 month ago
13

Imagine you are in a small boat on a small pond that has no inflow or outflow. If you take an anchor that was sitting on the flo

or of the boat and lower it over the side until it sits on the ground at the bottom of the pond, will the water level rise slightly, stay the same, or lower slightly?Two students, Ian and Owen, are discussing this. Ian says that the anchor will still displace just as much water when it is sitting on the bottom of the pond as it does when it is in the boat. After all, adding the anchor to the boat causes the water level in the lake to rise, and so would immersing the anchor in the pond. So Ian reasons that both displacements would be equal, and the lake level remains unchanged.
Physics
1 answer:
Keith_Richards [3.2K]1 month ago
3 0

Response:

The water level in the pond will decrease.

Explanation:

According to Archimedes' principle, an object floating displaces a volume of water equivalent to its weight, while an object resting on the bottom displaces an amount of water that corresponds to its volume.

When the anchor is aboard the boat, it acts as a floating object, and when it rests at the pond's bottom, it becomes a submerged object.

Due to the anchor's weight and density being greater than water, the amount of water displaced when it's in the boat exceeds the amount displaced when it is on the bottom of the pond since the anchor's volume is minimal.

Thus, once the anchor is dropped to the bottom, the pond's water level will decrease. If the anchor remains suspended, it continues to displace water as a floating body, causing no change, but once it contacts the pond's bed, the water level drops.

I hope this clarifies!

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Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.
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Consequently, we can ascertain the temperature rise \Delta T:
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Initially, the water's temperature was 25^{\circ}C, so the end temperature should be
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THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
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Answer: total angular distance = 1700° and 29.7 rad

   the total angular displacement = 0

Explanation:

This is a breakdown of the calculations needed.

The task is to determine both the total angular distance and displacement experienced by the knee.

To calculate the distance traveled by the knee, consider that when squatting, the knee bends 85° to lower, and then another 85° to return to standing (upright). Thus, the cumulative angular movement during the squat totals 170°.

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10 * 170° = 1700°

As such, the total angular distance reached is 1700°.

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Since 2π equals 360°, it follows that one degree equates to about 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1700° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

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