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Andrej [43]
4 years ago
9

The GPE of a 70kg person standing on a chair 1m off the ground is how many joules?

Physics
1 answer:
Artemon [7]4 years ago
8 0
GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity. 

U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J

U = 686.7J

Hope this helps!
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A force (5ỉ - ;)N moves an object from the point P (1,3) m to the point Q (3, 8) m.
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Answer:

b) 5 J

Explanation:

Work is the energy transferred by an object when acted by a force along a displacement. Work is the product of force and displacement. The SI unit of work is the joules (J)

To calculate the work done by the force, we have to first get the displacement (D) of the object. Hence:

Displacement (D) = Q(3, 8) - P(1, 3) = (3 - 1, 8 - 3) = (2, 5) = 2i + 5j

The work done is the dot product of the force and the displacement. Force = 5i - j. Hence:

Work done = (5i - j)(2i + 5j) = 10 - 5 = 5 J

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One end of a spring with a force constant of k = 10.0 N/m is attached to the end of a long horizontal frictionless track and the
koban [17]

Answer:

-2.478

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Explanation:

Angular frequency of spring in harmonic motion is given by?

ω = √(k/m)

ω = √(10/2.2)

ω = √4.54

ω = 2.13 s^-1

If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine

x(t) = -A * cos(ωt)

x(t) = -2.48 * cos(2.13 * 1)

x(t) = -2.48 * 0.9993

x(t) = -2.478

Velocity and acceleration are 1st and 2nd derivative of position

b)

v(t) = Aω * sin(ωt)

v(t) = 2.48 * 2.13 * sin(2.13 * 1)

v(t) = 5.282 * sin2.13

v(t) = 5.282 * 0.03717

v(t) = 0.379 m/s

c)

a(t) = Aω^2 * cos(ωt)

a(t) = 2.48 * 2.12² * cos(2.13 * 1)

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d)

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F = 24.78 N

3 0
3 years ago
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