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Ede4ka [16]
3 years ago
11

A loudspeaker in a parked car is producing sound whose frequency is 20 510 Hz. A healthy young person with normal hearing is sta

nding nearby on the sidewalk but cannot hear the sound because the frequency is too high. When the car is moving, however, this person can hear the sound. (a) Is the car moving toward or away from the person
Physics
1 answer:
Vsevolod [243]3 years ago
4 0

Answer:

The car must be moving away from the person.

Explanation:

From Doppler's Effect, we know that when a sound source moves towards a stationary observer, the apparent frequency of that sound increases. While the apparent frequency decreases if the source moves away from the stationary observer.

The audible range of frequencies for a human ear is 20 Hz to 20000 Hz. Therefore, in order for the sound of a loud speaker to be audible for the person, the frequency must decrease below 20000 Hz.

<u>Due to this reason, the car must be moving away from the person.</u>

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Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

6 0
3 years ago
Read 2 more answers
Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the
kykrilka [37]

Answer:

A) -2N

B) Left

C) -0.5

Explanation:

A) -12 + 10

B) More force is acted on in that direction

C) Net force/Mass (-2/4)

5 0
2 years ago
A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kg hangs from it. Th
den301095 [7]

Answer:

The function that describe the motion in the time

y (t) = 0.28m * sin ( 36.025 * t)

Explanation:

The angular frequency of oscillation of the spring

w = √k/m

w = √305 N/m / 0.235 kg

w = 36.025 rad / s

To determine the function of the motion knowing as a motion oscillation in a amplitude a frequency

y(t) = A * sin (w t )

So

A = 28.0 cm * 1 m / 100 cm = 0.28 m

So replacing to determine the function of the motion in the time

y (t) = A sin (w t)

y (t) = 0.28m * sin ( 36.025 * t)

7 0
3 years ago
A Martian rover found an interesting rock that could contain signs of what Mars was like long ago. It grabbed
kodGreya [7K]

Answer:

It would be PE=16kg * 9.8 m/s^2 * 1m = 160 J

Explanation:

The person who asked this question ended up answering his own question so I'm here to let you know all that the answer was founded by the person whos posted the question himself full credit goes to him :)

3 0
2 years ago
a crowbar of 2 meter is used to lift an object of 800N if the effort arm is 160cm , calculste the effort applied
Vitek1552 [10]

Answer:

200 N

Explanation:

The crowbar is 2 meter, or 200 cm.  The effort arm is 160 cm, so the moment arm of the object is 40 cm.

(800 N) (40 cm) = F (160 cm)

F = 200 N

5 0
3 years ago
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