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Yanka [14]
2 years ago
14

How long is a pendulum with a period of 1.0 s on a planet with twice the gravity of the earth?

Physics
1 answer:
UkoKoshka [18]2 years ago
6 0

Explanation:

T = 2π √(L / g)

1.0 s = 2π √(L / 19.6 m/s²)

L = 0.50 m

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Is the magnitude of the acceleration of the center of mass of spool A greater than, less than, or equal to the magnitude of the
Anon25 [30]

Answer:

The acceleration of the centre of mass of spool A is equal to the magnitude of the acceleration of the centre of mass of spool B.

Explanation:

From the image attached, the description from the complete question shows that the two spools are of equal masses (same weight due to same acceleration due to gravity), have the same inextensible wire with negligible mass is attached to both of them over a frictionless pulley; meaning that the tension in the wire is the same on both ends.

And for the acceleration of both spools, we mention the net force.

The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.

For this system, the net force on either spool is exactly the same in magnitude because the net force is a difference between the only two forces acting on the spools; the tension in the wire and their similar respective weights.

With the net force and mass, for each spool equal, from

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Hope this Helps!

6 0
2 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.200 rev/s . The magnitude
blagie [28]

Answer:

(A). the angular velocity of fan is 0.380 rev/s.

(B). The number of revolution is 0.059 rad.

(C). The tangential speed of the blade is 0.907 m/s.

(D). The tangential acceleration of the blade is 2.10 m/s²

Explanation:

Given that,

Initial angular velocity = 0.200 rev/s

Angular acceleration = 0.883 rev/s²

Diameter = 0.760 m

Time = 0.204 s

(A). We need to calculate the angular velocity of fan

Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value into the formula

\omega_{f}=0.200+0.883\times0.204

\omega_{f}=0.380\ rev/s

(B). We need to calculate the number of revolution

Using formula of angular displacement

\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2

Put the value into the formula

\theta=0.200\times0.204+\dfrac{1}{2}\times0.883\times(0.204)^2

\theta=0.059\ rad

(C). We need to calculate the tangential speed of the blade

Using formula of speed

v=\dfrac{d}{2}\omega_{f}

Put the value into the formula

v=\dfrac{0.760}{2}\times0.380\times2\pi

v=0.907\ m/s

(D). We need to calculate the tangential acceleration of the blade

Using formula of tangential acceleration

a_{t}=\alpha r

Put the value into the formula

a_{t}=0.883\times0.38\times2\pi

a_{t}=2.10\ m/s^2

Hence, (A). the angular velocity of fan is 0.380 rev/s.

(B). The number of revolution is 0.059 rad.

(C). The tangential speed of the blade is 0.907 m/s.

(D). The tangential acceleration of the blade is 2.10 m/s²

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