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Ksju [112]
3 years ago
5

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

it an additional 0.12 m
Physics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

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This component is called the piston.

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If the clearance is not sufficient the piston will seize and if it is excessive there will be loss of compression.

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3 years ago
Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow ________ about the center. a.counterclockwise an
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Answer:

counterclockwise and inward

Explanation:

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4 0
3 years ago
An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When
sleet_krkn [62]

Answer:

2dQ²/2εA = 2U₀

Explanation:

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U = Q²/2C = Q²/2εA/d = dQ²/2εA. At distance d₁ = d, U = U₀,

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3 0
3 years ago
A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/
Furkat [3]

Answer:

1) 23.45 rad/s²

2) 2.7 m/s²

3) t= 1.6 s

4) x ≈ 11 m

5) vfinal = 4.45 m/s

6) KErot = 16.2 J

    KEtran = 41 J

    KErot < KEtran

Explanation:

Step 1: Data given

mass bowling ball = 4.1 kg

radius = 0.117 meter

initial speed = 8.9 m/s

1) What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?

α = a / r = 2.774 m/s² / 0.117m = 23.45 rad/s²

2)What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?

a = µ*g = 0.28 * 9.8m/s² = 2.744 m/s² ≈ 2.7 m/s²

3) How long does it take the bowling ball to begin rolling without slipping?

This begins when ω = v / r

with

⇒ ω = α*t = 23.45 rad/s² * t

⇒ v = Vo - a*t = 8.9m/s - 2.744m/s²*t

This gives us:

23.45rad/s² * t = (8.9m/s - 2.744m/s²*t) / 0.11m

2.744*t = 8.9 - 2.744*t

t = 8.9 / 5.488 = 1.622 s ≈ 1.6 s

4) How far does the bowling ball slide before it begins to roll without slipping?

x = Vo*t - ½at² = (8.9*1.622 - ½*2.744*(1.622)²) m = 10.82 m ≈ 11 m

5) What is the magnitude of the final velocity?

v = Vo - at = 8.9m/s - 2.744m/s² * 1.622s = 4.45 m/s

6) After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball:

trans KE = ½ * 4.1kg * (4.45m/s)² =40.595 J ≈ 41 J

I = (2/5)mr² = (2/5) * 4.1kg * (0.117m)² = 0.0224 kg·m²

ω = v/r = 4.45m/s / 0.117m = 38.03 rad/s, so

rot KE = ½Iω² = ½ * 0.0224kg·m² * (38.03rad/s)² = 16.2 J

16.2 J < 41 J

KErot < KEtran

(For a rolling solid sphere, KErot ≈ 2/5 * KEtran)

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3 years ago
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Licemer1 [7]

Answer:

5016.2446

Explanation:

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