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Lubov Fominskaja [6]
2 years ago
7

What is the velocity of a jet that travels 785 m [W] in 2.39 s?

Physics
1 answer:
Cerrena [4.2K]2 years ago
7 0

Answer:

v = 328.4

Explanation:

Data:

  • Distance (d) = 785 m
  • Time (t) = 2.39 s
  • Velocity (v) = ?

Use formula:

  • \boxed{v = \frac{d}{t}}

Replace:

  • \boxed{v = \frac{785m}{2.39s}}

It divides:

  • \boxed{v = 328.4\frac{m}{s}}}

What is the velocity?

The velocity is <u>328.4 meters per second.</u>

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Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
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(b)
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Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
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(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
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In multiples of ten:
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(e)
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W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

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