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scoundrel [369]
3 years ago
9

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down,

5.00 s after it was thrown. What is the speed of the rock just before it reaches the water 22.0 m below the point where the rock left your hand? Ignore air resistance.
Physics
1 answer:
Agata [3.3K]3 years ago
7 0

Answer:

Explanation:

for vertical movement , time to reach the top = time to reach the hand = 2.5 s

v = u - gt

At the top , v = 0 , time t = 2.5 s

0 = u - g x 2.5

u = 2.5 x 9.8 = 24.5 m /s

velocity of throw = 24.5 m /s

So , when it passes the hand on its way down , it will have velocity equal to 24.5 m /s and it will accelerate downwards . Let its velocity down by 22 m be v

v² = u² + 2 g s

= 24.5² + 2 x 9.8 x 22

= 600.25 + 431.2

= 1031.45

v = 32.11 m /s .

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Vladimir [108]

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

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3 years ago
Playing with a stress ball you squeeze it as
shutvik [7]

Answer:

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Explanation:

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8 0
3 years ago
A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the
Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

<u>Given</u>:

           Initial angular velocity = ωi = 2.70 rad/s

           Final angular velocity = ωf = -2.70 rad/s (negative sign is  

           due to the movement in opposite direction)

           Change in time period = Δt = 1.50 s

<u>Required</u>:

           Change in angular velocity = Δω = ?

           Average angular acceleration = αav = ?

<u>Solution</u>:

          <u>Angular velocity (Δω):</u>

               Δω = ωf - ωi

               Δω = -2.70 - 2.70

               Δω = -5.4 rad/s.

          <u> Average angular acceleration (αav):</u>

               αav = Δω/Δt

               αav = -5.4/1.50

              αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0
3 years ago
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Rick is moving a wheelbarrow full of bricks out to the curb. The bricks in the wheelbarrow weigh more than Rick is able to carry
USPshnik [31]

Answer is given below

Explanation:

  • This is happen because here when Rick walks with full loaded wheelbarriow of bricks, he able to move it because Rick lifts the wheelbarrow handle
  • So, most of the weight of full loaded wheelbarrow's load goes on that's wheel and due to friction force between wheel and surface it can easy to move
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3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
Marizza181 [45]

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>
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