Yes, an object that was set in motion in the past by some force, but that is no longer being acted on by a net force, is moving but with zero acceleration, i.e. it is moving at constant velocity.

7 0

3 years ago

If an engine at sea level produces 100 horsepower, how many horsepower would it develop at 6,000 feet of altitude?

Sonja [21]

Answer:

Power develop at 6000 ft = 100 - 18 = 82 hp

Given:

Power produced by the engine at sea level = 100 hp

Explanation:

For each 1000 ft rise above the sea level, the power loss of the engine is about 3%

Therefore,

1000 ft - 3% loss

6000 ft - $3\times 6 = 18 %$

At 6000 ft, the power loss is 18% or 18 hp

Now,

Power develop at 6000 ft = Power at sea level - Power loss at 6000 ft

Power develop at 6000 ft = 100 - 18 = 82 hp

7 0

4 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

4 years ago