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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 rock is dropped from a garage roof from rest. the roof is 6.0 m from the ground. determine the velocity of the rock as it hits
Dmitrij [34]

From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity.  Velocity is a vector quantity.

it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

Given data:

V(Final velocity)=? (m/sec)

h(height)= 6.0 m

u(Initial velocity)=0 m/sec

g(gravitational acceleration)=9.81 m/s²

Newton's third equation of motion:

\rm v_y^2 = u_y^2+ 2gh \\\\\rm v_y^2 = 0+ 2gh\\\\\  v_y= \sqrt{2\times 9.81 \ (m/s^2)\times 6.0 (m)} \\\\ v_y=10.849 \ m/sec

Hence, the velocity of the rock as it hits the ground will be 10.849 m/sec.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

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A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying p
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Answer:5 N

Explanation:

Given

A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N

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N=\frac{0.200}{0.04}=5 N

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