A kayaker paddles with a power output of 50.0 W to maintain a steady speed of 1.30 m/s. Calculate the resistive force exerted by the water on the kayak. 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?

Since power = (resistive force) x (speed), the resistive force is 50 W/(1.3 m/s) = 38.5 N... which is about 8.5 pounds

If the resistive force stayed the same (which it won't), power would be proportional to speed.

BRUHH, HOW DOES A ROBOT HAVE A OIUTPUT WTF IS THIS CODING???? DONT YOU KNOW HOW TO CODE?? WHY SO SERIOUS. AN WE TALK ABOUT THE ECONOMICAL AND POLITICAL STATE OF THE WORK RN LIKE BRUIHHHH

To calculate the resistive force exerted by the water on the kayak, we can use the formula:

Resistive Force (Fr) = Power Output (P) / Speed (v)

Given that the power output (P) is 50.0 W and the speed (v) is 1.30 m/s, we can substitute these values into the formula to find the resistive force:

Fr = 50.0 W / 1.30 m/s
Fr ≈ 38.46 N

Therefore, the resistive force exerted by the water on the kayak is approximately 38.46 N.

Now, let's calculate the kayaker's new speed if she doubles her power output while the resistive force remains the same.

Initially, the power output is 50.0 W, and the speed is 1.30 m/s.
If the power output is doubled, it becomes 2 * 50.0 W = 100.0 W.

Using the same formula as before, we can find the new speed (v'):

Fr = P / v'
=> v' = P / Fr
=> v' = 100.0 W / 38.46 N
v' ≈ 2.60 m/s

Therefore, if the kayaker doubles her power output while the resistive force remains the same, her speed will increase to approximately 2.60 m/s. The kayaker's speed is multiplied by a factor of 2.60 / 1.30 = 2.

Well the kayaker is probably a robot if it has an exact power output, so just change the settings to measure the speed of the water that's passing.

two times.