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3 years ago

A man pushes a lawn mower on a level lawn with a force of 186.5 N. If 3.8% of this

Physics

1 answer:

Paladinen [302]3 years ago

4 0

Explanation:

The force is 186.5 N. 3.8% of the force is directed downward. So the vertical component of the force is:

Fᵧ = 0.038 (186.5 N)

Fᵧ = 7.087 N

The horizontal component is found with Pythagorean theorem:

F² = Fₓ² + Fᵧ²

(186.5 N)² = Fₓ² + (7.087 N)²

Fₓ = 186.4 N

So the work done is:

W = Fd

W = (186.4 N) (7.4 m)

W = 1379 J

Round as needed.

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3 years ago

What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction

Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

$f'=f(\dfrac{v+v_{p}}{v-v_{s}})$

Where, f = frequency

v = speed of sound

$v_{p}$ = speed of passenger

$v_{s}$ = speed of source

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$f'=262\times(\dfrac{344+18}{344-30})$

$f'=302.05\ Hz$

Hence, The frequency is 302.05 Hz.

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3 years ago

To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit

DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

$R = 350ft$

$a_t = 1.1ft/s^2$

PART A )

$a_c = \frac{V^2}{R}$

$a_c = \frac{V^2}{350}$

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $5.25ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$27.5625 = 1.21 + \frac{v^4}{122500}$

$v=42.3877ft/s$

Now calculate the angular velocity of the motorcycle

$v = r\omega$

$42.3877 = 350\omega$

$\omega = 0.1211rad/s$

Calculate the angular acceleration of the motorcycle

$a_t = r\alpha$

$1.1 = 350\alpha$

$\alpha = 3.1428*10^{-3}rad/s^2$

Calculate the time needed by the motorcycle to reach an acceleration of

$5.25ft/s^2$

$\omega = \alpha t$

$0.1211 = 3.1428*10^{-3}t$

$t = 38.53s$

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

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PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

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$a_r = \frac{21.5^2}{350}$

$a_r =1.3207ft/s^2$

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

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$a = \sqrt{(1.1)^2+(1.3207)^2}$

$a = 1.7187ft/s^2$

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$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

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3 years ago

what happens to the current across a circuit when the voltage is doubled while the resistance is held back

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Since the current is inversely propotional to its resistance, when the voltage is doubled the current will be one-half

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Explanation:

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3 years ago