Answer:
45000 atoms is the correct answer.
Explanation:
Answer: Negative acceleration
Explanation:
According to the described situation we have two velocities:
-An initial velocity during the first hour:
-A final velocity during the last 200 miles:
As we can see, the final velocity is less than the initial velocity, this means the plane's velocity decreased.
On the other hand, acceleration is defined as <u>the variation of velocity in time:</u>
Since the plane's velocity decreases, the acceleration is negative.
Hence, this situation is an example of negative acceleration.
Answer:
0.70 s
Explanation:
Potential energy = kinetic energy + rotational energy
mgh = ½ mv² + ½ Iω²
For a thin spherical shell, I = ⅔ mr².
mgh = ½ mv² + ½ (⅔ mr²) ω²
mgh = ½ mv² + ⅓ mr²ω²
For rolling without slipping, v = ωr.
mgh = ½ mv² + ⅓ mv²
mgh = ⅚ mv²
gh = ⅚ v²
v = √(1.2gh)
v = √(1.2 × 9.81 m/s² × 1.1 m sin 49.0°)
v = 3.13 m/s
The acceleration down the incline is constant, so given:
Δx = 1.1 m
v₀ = 0 m/s
v = 3.13 m/s
Find: t
Δx = ½ (v + v₀) t
t = 2Δx / (v + v₀)
t = 2 (1.1 m) / (3.13 m/s + 0 m/s)
t = 0.704 s
Rounding to two significant figures, it takes 0.70 seconds.
Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v