take the difference of water levels to find volume of the ring. convert ml to cm3. multiply by density to cancel out cm3 units (idk how to explain that well). you get 8.4g
Answer:
in t seconds, Car A sweep out t radian { i.e θ = t radian }
Explanation:
Given the data in the question;
4 toy racecars are racing along a circular race track.
They all start at 3 o'clock position and moved CCW
Car A is constantly 2 feet from the center of the race track and moves at a constant speed
so maximum distance from the center = 2 ft
The angle Car A sweeps out increases at a constant rate of 1 radian per second.
Rate of change of angle = dθ/dt = 1
Now,
since dθ/dt = 1
Hence θ = t + C
where C is the constant of integration
so at t = 0, θ = 0, the value of C will be 0.
Hence, θ = t radian
Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }
Assuming the woman starts at rest, she descends the slide with acceleration <em>a</em> such that
(20.3 m/s)² = 2 <em>a</em> (42.6 m) → <em>a</em> ≈ 4.84 m/s²
which points parallel to the slide.
The only forces acting on her, parallel to the slide, are
• the parallel component of her weight, <em>w</em> (//)
• friction, <em>f</em>, opposing her descent and pointing up the slide
Take the downward sliding direction to be positive. By Newton's second law, the net force in the parallel direction acting on the woman is
∑ <em>F</em> (//) = <em>w </em>(//) - <em>f</em> = <em>ma</em>
where <em>m</em> = 77.0 kg is the woman's mass.
Solve for <em>f</em> :
<em>mg</em> sin(42.3°) - <em>f</em> = <em>ma</em>
<em>f</em> = <em>m</em> (<em>g</em> sin(42.3°) - <em>a</em>)
<em>f</em> = (77.0 kg) ((9.80 m/s²) sin(42.3°) - 4.84 m/s²) ≈ 135 N
Compute the work <em>W</em> done by friction: multiply the magnitude of the friction by the length of the slide.
<em>W</em> = (135 N) (42.6 m) ≈ 5770 N•m = 5770 J
Answer:
200 meters+1000 meters+800 meters=2000 meters or 2km
Explanation:
1km=1000m