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dybincka [34]
3 years ago
15

Jasper and Gemma are going to play on a teeter totter. Gemma gets on first. When Jasper gets on, Gemma moves into the air, but s

he does not move to the top.
Which statement could correctly explain the forces acting on the teeter totter? assume that Jasper and Gemma are the same distance from one another.

The forces are balanced because Jasper weighs the same as Gemma.
The forces are unbalanced because Jasper weighs the same as Gemma.
The forces are balanced because Jasper weighs more than Gemma.
The forces are unbalanced because Jasper weighs more than Gemma.
Physics
2 answers:
Nezavi [6.7K]3 years ago
7 0

Answer:

it will most likely be a: the forces are balanced because jasper weighs the same as gemma.

Explanation:

Aleksandr [31]3 years ago
5 0

Answer:

A

Explanation:

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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
2 years ago
While standing on a balcony a child drops a penny. The penny lands on the ground floor 1.5 s later. How fast was the penny trave
sveticcg [70]

Answer:

14.7 m/s.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 1.5 s

Acceleration due to gravity (g) = 9.8 m/s².

Height = 11.025 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

We, can obtain the initial velocity of the penny as follow:

H = ½(v + u) t

11.025 = ½ (0 + u) × 1.5

11.025 = ½ × u × 1.5

11.025 = u × 0.75

Divide both side by 0.75

u = 11.025/0.75

u = 14.7 m/s

Therefore, the penny was travelling at 14.7 m/s before hitting the ground.

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