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nlexa [21]
1 year ago
12

The position of an object that is oscillating on a spring is given by the equation x = (17.4 cm) cos[(5.46 s-1)t]. what is the a

ngular frequency for this motion?
Physics
1 answer:
nekit [7.7K]1 year ago
8 0

the angular frequency of this motion is 5.46rad /sec

The formula for the angular frequency is = 2π/T. Radians per second are used to express angular frequency. The frequency, f = 1/T, is the period's inverse. The motion's frequency, f = 1/T = ω/2π, determines how many complete oscillations occur in a given amount of time so in this case the It is measured in units of Hertz, (1 Hz = 1/s).

herex

=

(17.4cm)cos[(5.46s− 1)t]

is written in the general form

where we can identify: A=17.4cm and ω

=5.46rad /sec

To learn more about angular frequency :

brainly.com/question/12446100

#SPJ4

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If two charged balloons are 24cm apart and they feel a force of electrical repulsion of 20N, what would the force of electrical
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5 0
3 years ago
The mass of Jupiter is 1.9 × 1027 and that of the sun is 1.99 × 1030. The
n200080 [17]

Answer:

F = 4.147 × 10^23

v = 1.31 × 10^4

Explanation:

Given the following :

mass of Jupiter (m1) = 1.9 × 10^27

Mass of sun (m2) = 1.99 × 10^30

Distance between sun and jupiter (r) = 7.8 × 10^11m

Gravitational force (F) :

(Gm1m2) / r^2

Where ; G = 6.673×10^-11 ( Gravitational constant)

F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2

F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)

F = (25.231 × 10^46) / (60.84 × 10^22)

F = 3.235 × 10^(46 - 22)

F = 0.4147 × 10^24

F = 4.147 × 10^23

Speed of Jupiter (v) :

v = √(Fr) / m1

v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)

v = √32.3466 × 10^(23+11) / 1.9 × 10^27

v = √32.3466× 10^34 / 1.9 × 10^27

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v = √17.023 × 10^7

v = 13047.221

v = 1.31 × 10^4

4 0
3 years ago
A box sits on a table. A short arrow labeled F subscript N points up. A short arrow labeled F subscript g points down. A long ar
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unbalanced and right

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7 0
3 years ago
If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?
Lunna [17]

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on  the skier is,

   W=T.S

⇒W=270*160*cos\pi \\W=-43200 J

Hence work done by the rope is - 43200 J.

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8 0
2 years ago
Read 2 more answers
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