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sergiy2304 [10]
3 years ago
10

Can carrying an object be said to be doing work on that object?

Physics
2 answers:
sesenic [268]3 years ago
8 0
Only if that object is being moved by carrying
Schach [20]3 years ago
7 0

Answer:

Only if that object is being moved while carried

Explanation:

Work = Force • distance travelled

While holding an object would need force equal to that objects mass multiplied by 9.8, if it hasn't moved, then no work is done.

if you were to carry the object and have it travel some distance, then work would be done on that object.

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A car is traveling at a speed of 15m/s. It experiences an acceleration of -3m/s^2, that lasts for 4 seconds. What is the final v
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Answer:

27m/s

Explanation:

u=15m/s

a=3m/s^2

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v=u+at

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A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. T
lukranit [14]

Answer:

(a) The period is 4s

(b) The angular frequency is pi/2 radians

(c) The amplitude is 0.37cm.

(d) The displacement at time is  (0.37 cm) cos((pi/2)*t)

(e) The Velocity at time t is v = (0.58 cm)(sin((pi/2)*t)

(f) The maximum speed is  v_{m} = -0.58 cm/s

(g) The  maximum acceleration is 0.91 cm/s^2

Explanation:

We have a particle which oscillates with frequency of f = 0.25 Hz about the point x = 0.At t = 0, the displacement of the particle is = 0.37 cm and its velocity is zero.

(a) The period of the oscillations is,

T = 1/f

so

T = 1/(0.25 Hz)

T = 4.0s

(b) The angular frequency is,

f = 2\pi f\\

f = = 2(\pi)(0.25 Hz) =\pi /2  \\ radians

(c) Since

The amplitude is the maximum displacement that the particle makes from the equilibrium point, or when the speed of the particle is zero,

that is

x_{m}= 0.37 cm

(d) The displacement as a function of t is given be,

x = x_{m} cos(ωt+Φ)

as x = x_{m  t = 0, we get cos(Ф) = 1 = 0

so this equation becomes

x= (0.37 cm) cos((pi/2)*t)

(e) Now we need to find the speed of the particle as a function of t

the speed is the derivative of the displacement that is

v = dx/dt = -(0.37)(pi/2)(sin((pi/2)*t)

so the velocity at time t is

 v = (0.58 cm)(sin((pi/2)*t)

(f) Since

v = v_{m} sin(ωt+Ф)

then from part (e) we get

v_{m} = -0.58 cm/s

(g)

The amplitude of the maximum acceleration is

a_{m} = x_{m ω^2

      = (0.37 cm) (pi/2) = 0.91 cm/s^2

this is the maximum acceleration

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3 years ago
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