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1 year ago

What is the speed of a truck traveling 10km in 10 minutes

Physics

1 answer:

user100 [1]1 year ago

8 0

Answer: 58.8235 km/h

speed = distance/time

the distance is 10 km

the time is 10 minutes

the unit is not correct, so we first change minute to hour

so 10/60 is 0.166667, rounded to 0.17.

10 km/ 0.17 hours =

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3 years ago

Newton's second law is a major part of mechanics. However, it does have its limitations. Under what condition does the second la

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3 years ago

En una práctica un beisbolista lanza verticalmente la bola con una velocidad de 12m/s en dirección ascendente. ¿Cuál será la alt

fgiga [73]

Answer:

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3 years ago

A child of mass M is swinging on a swing set. The ropes attaching the swing to the top bar have length L. Find the gravitational

myrzilka [38]

Answer:

(a) 0

(b) 10ML

(d) $10ML(1 + sin(\phi))$

Explanation:

(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.

(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be

$E_h = mgh = 10ML$

where g = 10m/s2 is the gravitational acceleration.

(c) At angle $\theta$ from the vertical. Vertically speaking, the child should be at a distance of $Lcos(\theta)$ to the swinging point, and a vertical distance of $L - Lcos(\theta)$ to the lowest position. His potential energy to this point would be:

$E_{\theta} = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))$

(d) at angle $\phi$ from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of $Lsin(\phi)$ from the swinging point and higher than it. Therefore his vertical distance to the lowest point is $L + Lsin(\phi) = L(1 + sin(\phi))$