The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of
950 N = (2300 kg) <em>a</em>
<em>a</em> = (950 N) / (2300 kg)
<em>a</em> ≈ 0.413 m/s²
Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of
<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)
<em>v</em> = -0.75 m/s
which means the wind slows the boat down to a velocity of 0.75 m/s westward.
Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before.
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.
This formula tells how far an object travels in how much time,
when it's accelerating:
Distance = (1/2 acceleration) x (Time²).
D = 1/2 A T²
For your student who dropped an object out of the window,
Distance = 19.6 m
Acceleration = gravity = 9.8 m/s²
D = 1/2 G T²
19.6 = 4.9 T²
Divide each side by 4.9 : 4 = T²
Square root each side: 2 = T
When an object is dropped in Earth gravity,
it takes 2 seconds to fall the first 19.6 meters.
45mph is the answer if you do the math right
Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity 
= 60 mph = 26.8224 m/s
Final velocity 
= 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = 
m( ² - ² )
we substitute
Δk = ×1814.37( (26.8224)² - (13.4112)² )
Δk = × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
