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77julia77 [94]
3 years ago
7

HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!​

Physics
2 answers:
kicyunya [14]3 years ago
8 0

Answer:

25000 J work with 2500 W power

Explanation:

work done = force . displacement = 500 N . 50 m= 25000 J

Power= work done / time taken = 25000 J / 10 s = 2500 W

myrzilka [38]3 years ago
4 0

Answer:

a. 25000J or 25KJ

b.2500w

Explanation:

a. since we say work is force applied in the same line or straight line

<em>W=</em><em>F(</em><em>newtons </em><em>)</em><em>*</em><em>D(</em><em>metres</em><em>)</em><em> </em>

<em>b.</em><em> </em><em>power </em><em>is </em><em>work </em><em>over </em><em>time </em>

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Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

v_{f}^{2} = v_{i}^{2} + 2ad        

v_{f} = \sqrt{v_{i}^{2} + 2ad}  (1)

The acceleration can be found by means of Newton's second law:

\sum F_{net} = ma

Where \sum F_{net} is the net force, m is the mass and a is the acceleration.

Fx + Fy = ma  (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

F_{x} = F - F_{air}  (3)

Forces in the y axis:

F_{y} = 0 (4)

Solving for the forces in the x axis:

F_{x} = F - F_{air}

Where F = 1.150x10^{3} N and F_{air} = 795 N:

F_{x} = 1.150x10^{3} N - 795 N

F_{x} = 355 N

Replacing in equation (2) it is gotten:

Fx + Fy = ma

355 N + 0 N = (90.0 Kg)a

355 N = (90.0 Kg)a

a = \frac{355 N}{90.0Kg}

a = \frac{355 Kg.m/s^{2}}{90.0Kg}

a = 3.94 m/s^{2}

So the acceleration for the cyclist is 3.94 m/s^{2}, now that the acceleration is known, equation (1) can be used:

v_{f} = \sqrt{v_{i}^{2} + 2ad}

However, since he was originally at rest its initial velocity will be zero (v_{i} = 0).

v_{f} = \sqrt{2ad}

v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}

v_{f} = 16.60m/s

He has a speed of 16.60m/s after 35.0 meters

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Your answer is C. Unbalanced

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To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

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I = \frac{100}{4\pi (2.5)^2}

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