contestada

A kayaker paddles with a power output of 60.0 W to maintain a steady speed of 1.40 m/s a.)Calculate the resistive force exerted by the water on the kayak. b.) If the kayaker doubles her power output, and the resistive force due to the water remains the same, by what factor does the kayaker's speed change?

Respuesta :

Answer:

a)F= 42.85 N

b)v=2.8 m/s

Explanation:

Given that

Power ,P= 60 W

Speed,v= 1.4 m/s

We know that power P given

P = F .v

F=force

v=speed

P=Power

a)

P = F .v

60 = F  x 1.4

[tex]F=\dfrac{60}{1.4}\ N[/tex]

F= 42.85 N

Therefore the force will be 42.85 N

b)

If the power become 120 W and force is constant 428.57 N

The speed = v m/s

P = F .v

120 = 428.57  x v

[tex]v=\dfrac{120}{42.85}\ N[/tex]

v=2.8 m/s

Therefore the new speed will be 2.8 m/s.

Cricetus

(a) The force will be "42.85 N".

(b) The new speed will be "2.8 m/s".

Given:

  • Power, P = 60 W
  • Speed, v = 1.4 m/s

(a)

As we know the formula,

→ [tex]P = F\times v[/tex]

By putting the values, we get

  [tex]60 = F\times 1.4[/tex]

   [tex]F = \frac{60}{1.4}[/tex]

       [tex]= 42.85 \ N[/tex]

(b)

The speed:

→ [tex]P = F\times v[/tex]

[tex]120 = 428.57\times v[/tex]

   [tex]v = \frac{120}{42.85}[/tex]

      [tex]= 2.8 \ m/s[/tex]

Thus the above responses are correct.  

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