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Alexandra [31]
1 year ago
13

Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).

What note will you actually hear as you approach? (Assume a speed of sound of 340 m/s.)
Physics
1 answer:
MrRissso [65]1 year ago
7 0

Answer:

You will hear the note E₆

Explanation:

We know that:

Your speed = 88m/s

Original frequency = 1,046 Hz

Sound speed = 340 m/s

The Doppler effect says that:

f' = \frac{v \pm v0 }{v \mp vs}*f

Where:

f = original frequency

f' = new frequency

v = velocity of the sound wave

v0 = your velocity

vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

f' = \frac{340 m/s + 88m/s}{340 m/s} *1,046 Hz = 1,316.73 Hz

This frequency is close to the note E₆ (1,318.5 Hz)

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A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

\theta= \theta_0 +\omega t

From our given data we know that,

\omega = 76\frac{rev}{min}

\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

\theta_0 = 0.47 rad

Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

\theta= 0.47+(7.958)(8.1)

\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

\theta_{net} = 0.108*2\pi=0.216\pi

Therefore the angle of the speck at a time 8.1s is 0.216\pi

4 0
2 years ago
A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = \frac{4}{3} \pi (\frac{D}{2})^3

Therefore \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

r of the disc = 1.15(\frac{ D}{2} )

Using conservation of angular momentum;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disc = m*\frac{   \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{  m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
2 years ago
A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th
alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

F=mg

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

a_y = g = -9.8 m/s^2

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

v_x = v_0 cos \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s

D) 9.3 m/s

The y-component of the shot's velocity is given by:

v_y = v_0 sin \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

v_x = 7.6 m/s

F) -11.1 m/s

The y-component of the velocity at time t is given by:

v_y(t) = v_y + at

where

v_y = 9.3 m/s is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s

where the negative sign means the vertical velocity is now downward.

3 0
2 years ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




7 0
2 years ago
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BlackZzzverrR [31]

Answer:

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Explanation:

The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.

Or

If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.  

On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.

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