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timofeeve [1]
2 years ago
5

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

. Assume each tick mark represents 1 cm. Calculate the total displacement if a toy car starts at 0, moves 5 cm to the left, then 8 cm to the right, and then 3 cm to the left. The car moves cm, so there is no displacement.
Physics
2 answers:
Natali5045456 [20]2 years ago
6 0

The total displacement of the toy car at the given positions is 0.

The given parameters;

  • <em>First displacement of the car, = 5 cm left</em>
  • <em>Second displacement of the car, = 8 cm right</em>
  • <em>Third displacement of the car, = 3 cm to the left</em>

The total displacement of the car is calculated as follows;

  • <em>Let the </em><em>left </em><em>direction be "</em><em>negative </em><em>direction"</em>
  • <em>Let the </em><em>right </em><em>direction be "</em><em>positive </em><em>direction"</em>

\Delta x = - \ 5\ cm  \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0

Thus, the total displacement of the toy car at the given positions is 0.

Learn more about displacement here: brainly.com/question/18158577

jeka57 [31]2 years ago
5 0

Answer:0

Explanation:

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complete question:

An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°. What is the distance from the base of the cliff to the point on the ground? Round to the nearest foot

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a ≈  5281 ft

Explanation:

The observer at the top of a 462 ft cliff  measures the angle of depression from the top of the cliff to a point on the ground to be 5°.

The angle of depression form the top of the cliff = 5°

The 5° is outside the triangle formed . To find the angle in the triangle we have to subtract 5° from 90°.  90° - 5° = 85° Note sum of an angle on a right angle is 90°.  

using SOHCAHTOA  principle we can solve for the distance from the base of the cliff to the point on the ground(a)

tan  85° = opposite / adjacent

tan 85°  = a / 462

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462 × tan 85° = a

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Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks
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The concept of this question can be well understood by listing out the parameters given.

  • The mass of the block = 51 g = 51 × 10⁻³ kg
  • The distance of the block from the table = 30 cm
  • Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

\mathsf{S = ut + \dfrac{1}{2}gt^2}

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

  • Since 1 cm = 0.01 m
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The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}

\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}

\mathsf{0.3 =4.9*(t^2)}

By dividing both sides by 4.9, we have:

\mathsf{t^2 = \dfrac{0.3}{4.9}}

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\mathsf{t = \sqrt{0.0612}}

\mathsf{t =0.247  \ seconds}

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}

where the relation between time (t) and period (T) is:

\mathsf{t = \dfrac{T}{2}}

T = 2t

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\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\  \mathsf{  k = \dfrac{2 \pi^2*m}{T^2}}

\mathsf{  k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}

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Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

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