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MAVERICK [17]
3 years ago
7

A playground tire swing has a period of 2.0 s on Earth. What is the length of its chain?

Physics
1 answer:
Jet001 [13]3 years ago
5 0

Answer:

Length of the chain, l = 0.99 m

Explanation:

Given that,

A playground tire swing has a period of 2.0 s on Earth i.e.

T = 2 s

We need to find the length of this chain. The relationship between the length and the time period is given by :

T=2\pi \sqrt{\dfrac{l}{g}}

Where

l = length of chain

g = acceleration due to gravity

l=\dfrac{T^2g}{4\pi^2}

l=\dfrac{(2)^2\times 9.8}{4\pi^2}

l = 0.99 meters

So, the length of the chain is 0.99 meters. Hence, this is the required solution.

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3 years ago
A wave created by shaking a rope up and down is called a.
Sidana [21]
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5 0
1 year ago
A bus travels north on some busy city streets for 2.5 km, and a trip
d1i1m1o1n [39]

Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

                                                                                     = 2500 m

The time taken by the trip is, t = 9 min

                                                  = 540 s

The velocity of the bus,

                                       V = d / t

                                           = 2500 / 540

                                          = 4.63 m/s

At another  point, the bus travels at a constant speed of v = 18 m/s

Therefore the velocity becomes

                                                V = (4.63 + 18)/2

                                                   = 11.31 m/s

Hence, the velocity of the  bus, V = 11.31 m/s

8 0
3 years ago
The distance between two corresponding parts of a wave is its
oksano4ka [1.4K]

Answer:

wavelenght

Explanation:

The wavelength is the spatial period of a wave, analogous to the temporal period, it is the distance between two consecutive points with maximum amplitude that are repeated in space . In the waves of the sea, the wavelength is easily observed in the separation between two consecutive ridges.

4 0
3 years ago
Read 2 more answers
A bug is sitting on the edge of a rotating disk. At what angular velocity will the bug slide off the disk if its radius is 0.241
Ivahew [28]

Answer:

ω = 3.61 rad/sec

Explanation:

Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.

μmg = mv^2/r = mω^2r

Thus;

μg = ω^2r

ω^2 = μg/r

ω = √(μg/r)

ω = √(0.321 * 9.8)/0.241

ω = √(13.05)

= 3.61 rad/sec

3 0
3 years ago
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