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Komok [63]
3 years ago
12

Imagine that person B is more massive than person A in the picture above.

Physics
1 answer:
Masja [62]3 years ago
6 0
Hippity hoppity your points are my property
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A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 m from the axis of rotation.
dezoksy [38]

Hi there!

We can use the conservation of angular momentum to solve.

L_i = L_f\\\\I\omega_i = I\omega_f

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Recall the following equations for the moment of inertia.

\text{Solid cylinder:} I = \frac{1}{2}MR^2\\\\\text{Object around center:} = MR^2

Begin by converting rev/sec to rad sec:


\frac{0.17rev}{s} * \frac{2\pi rad}{1 rev} = 1.068 \frac{rad}{s}

According to the above and the given information, we can write an equation and solve for ωf.

1.068(\frac{1}{2}(34)(1.6)^2 + (79)(1.6)^2) = \omega_f(\frac{1}{2}(34)(1.6^2) + 79(0^2))\\\\\omega_f = \boxed{6.03 \frac{rad}{sec}}

4 0
3 years ago
Many compounds of living things, such as carbon dioxide, are formed by which kind of bond?
Anuta_ua [19.1K]

Answer:

im sorry i dont know

Explanation:

8 0
3 years ago
Read 2 more answers
Why does the light from the touch reach upto only certain distance
ad-work [718]
It is also likely (but not certain) that the photons will be absorbed by atoms. ... Light particles( or photons) never”run out” or loose their energy, so they can go an infinite distance, or until it reaches an object, that reflects the light or obsorbs it. Ie, a planet, or a mirror.
6 0
4 years ago
a 64kg skateboarder on a 2.0kg skateboard is on top of a ramp with a vertical height of 5.0 m what is the skateboarders maximum
WITCHER [35]
The formula for a kinetic energy KE of a falling body is
KE = mgh
where m = mass, g = acceleration due to gravity (9.8 m/s^2, constant), h = height.

The total mass of a skateboader and a skateboard is 64 + 2.0 = 66 kg.

Finally,
KE = 66*9.8*5.0 = 32340 J
7 0
3 years ago
The 68-kg crate is stationary when the force P is applied. Determine the resulting acceleration of the crate if (a) P = 0, (b) P
bogdanovich [222]

Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

a=\dfrac{P}{m}

a=\dfrac{181\ N}{68\ kg}

a=2.67\ m/s^2

(c) P = 352 N

a=\dfrac{P}{m}

a=\dfrac{352\ N}{68\ kg}

a=5.17\ m/s^2

Hence, this is the required solution.

5 0
3 years ago
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