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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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A mass of 4 Kg rest on the horizontal plane. The plane is gradually inclined until at an angle of θ=15

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2 years ago
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The pitching speed of your friend is 33.20 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

Your friend throw the ball horizontally that means the vertical initial

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<u><em>Remember:</em></u> the height is negative value because its below the point of

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h = -4 m , u_{y}=0 and g = -9.8 m/s²(downward)

<em>Substitute these values in the rule above</em>

⇒ 4=0-\frac{1}{2}(9.8)t^{2}

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ <em>t = 0.9035</em>

Then the time of the ball to land on the ground is 0.9035 seconds

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4 0
3 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

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E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

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q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

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F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

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Given data;

m(mass of sled)=8 kg

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To learn more about the force refer to the link;

brainly.com/question/26115859

#SPJ1

5 0
1 year ago
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