A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?

answer:

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°.

For 10 squats, the knee must undergo 170° motion multiplied by 10, resulting in:

10 * 170° = 1700°

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

Now converting this to radians since both degree and radian outputs are required:

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

For the second part, remember that angular displacement is determined by the angular distance divided by time, leading to a displacement of zero because the knee's ending position is the same as its starting position.

I hope this is helpful!!!!

π

Answer:

The distinction between the locations of the two events as perceived in S is calculated to be 4.42 x 10^8 m

Explanation:

The time dilation concept has been utilized to address this question, as detailed in the attachment.

Answer:

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The sheet's thickness is calculated to be t= 0.0003 mm. The mass provided is m = 230 mg, equivalent to 0.23 g. The area of the sheet is A= 23 x 17 cm², totaling A= 391 cm². Given the density of gold is ρ = 19.32 g/cm³, we assume the sheet's thickness is t cm. From the equation Mass = Density x Volume, we know m = ρ A t. Substituting the values results in: 0.23 = 19.32 × 391 × t, leading to t = 0.000030 cm, or equivalently, 0.0003 mm as the final thickness.