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ValentinkaMS [17]
2 years ago
9

Before working this problem, review Conceptual Example 14. A pellet gun is fired straight downward from the edge of a cliff that

is 14.7 m above the ground. The pellet strikes the ground with a speed of 27.2 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward
Physics
1 answer:
victus00 [196]2 years ago
5 0

Answer:

23.04 m

Explanation:

We'll begin by calculating the initial velocity of the pellet. This can be obtained as follow:

Height (h) of cliff = 14.7 m

Final velocity (v) = 27.2 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v² = u² + 2gh

27.2² = u² + (2 × 9.8 × 14.7)

739.84 = u² + 288.12

Collect like terms

u² = 739.84 – 288.12

u² = 451.72

Take the square root of both side

u = √451.72

u = 21.25 m/s

Thus, the initial velocity of the pellet is 21.25 m/s.

Finally, we shall determine the maximum height to which the pellet would have gone assuming the gun was fired straight upward. This can be obtained as follow:

Initial velocity (u) = 21.25 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

Maximum height (h) =?

v² = u² – 2gh (since the pellet is going against gravity.

0² = 21.25² – (2 × 9.8 × h)

0 = 451.5625 – 19.6h

Collect like terms

0 – 451.5625 = –19.6h

–451.5625 = –19.6h

Divide both side by –19.6

h = –451.5625 / –19.6

h = 23.04 m

Therefore, the pellet will reach a maximum height of 23.04 m above the cliff.

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(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

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A = 3.5 cm

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-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

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-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

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6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

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b = 0.93

Thus, the damping coefficient is 0.93.

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2 years ago
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The net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N. Details about net force can be found below.

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The net force of a body can be calculated by multiplying the mass of the body by its acceleration as follows:

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According to this question, a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right.

Net force = 8kg × 4m/s²

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Therefore, the net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N.

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E = P * t

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E = qV

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C = q / V

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P = V^2 / R

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Answer:

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A train going at a speed of 18m / s brakes and stops in 15s calculates its acceleration and the distance traveled when braking

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