The change in pitch of a train's horn as it passes while you are standing still can be explained by A. the Doppler effect.
B. ultrasonic vibrations.
C. sympathetic vibration.
D. resonance.

Answers

Answer 1

Answer: The change in pitch of a train's horn as it passes while you are standing still can be explained by the Doppler effect. The correct option among all the options given in the question is option "A". When the sound source from the horn of the train comes closer, the sound seems to come closer each time a sound wave gets emitted. So it comes closer to us in a bunched form. The reverse happens when the train has just passed with each emitted sound going further and further away. The frequency of the sound seems to get lower as the train moves further away.

Answer 2

Answer:

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Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the bottom of the first hill. You estimate that the efficiency of the tracks and cars you are using 90.0%. How high must the first hill be?

Answers

Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

   KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:

   KE = 0.9 PE

(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)

Divide each side by (mass):

   (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

   (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

   Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

      = (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

      = (24,500 / 88.2) (m²/s²) / (m/s²)

      = 277-7/9 meters
      (about 911 feet)

What force keeps a water spider on the surface of water

Answers

i think it's surface tension force

Money is useful to people because it is __________.Select one of the options below as your answer:a. a medium of exchange
b. prestigious
c. nice to look at
d. something that makes you feel good

Answers

a. A medium of exchange ....Hope am right !!!

The answer is A because money is a form of barter in a sense.

When you approach a light source, the wavelength of emitted light appears _____.a. shorter
b. the same
c. longer

Answers

When you approach a light source, or when the light source
approaches you, the wavelength of the light you see is shorter
than the wavelength of the light that's actually emitted by the source.

Similarly, when you approach a sound source, or when the sound
source approaches you, the wavelength of the sound you hear is
shorter than the wavelength of the sound that's actually emitted by
the source.

Answer:

A. shorter

Explanation:

I just took the test.

A clever engineer designs a "sprong" that obeys the force law fx =−q(x−xe)3, where xe is the equilibrium position of the end of the sprong and q is the sprong constant. for simplicity, we'll let xe =0m. then fx =−qx3.

Answers

The unit of q is N/m³.While the potential energy of the compressed spring will be $\rm U(x)= -(qx^4)/(4)$ .

What is the potential energy of spring?

When a springdeviates from its mean position, it attempts to regain equilibrium by applying a force that is equivalent to but opposite to the external force.

Spring force has been used in bicycle carriers and launchingmechanisms, where the energy generated by disrupting the spring'sbalance is used as potential energy.

The given equation for the force law is;

$\rm f_x =-q(x-x_e)$

Where,

$\rm x_e$ is equilibrium position of the end of the spring

q is a spring constant

If at equilibriumposition;

$\rm x_e=0$

$\rm F_X= -qx^3 \n\n \rm q = (F_x)/(x^3) \n\n$

So the unit derived will be;

q = N/m³

The potential energy of the spring is found by;

$\rm F_x=- (du)/(dx) \n\n U(x)= -\int\limits^x_0 {F(x)} \, dx \n\n U(x)= -\int\limits^x_0 -qx^3 \, dx \n\n U(x)=- (qx^4)/(4)$

Hence the unit of q is N/m³.While the potentialenergy of the compressed spring will be $\rm U(x)= -(qx^4)/(4)$ .

To learn more about the potential energy of spring refer to the link;

brainly.com/question/2730954

The objective of this question is to compute the units of q and calculate the expression for the P.E of the compressed spring

Answer:

q = N/m³

$\mathbf{U(x) = -(qx^4)/(4)}$

Explanation:

Given that:

The equation for the force law is:

$F_x = -q (x -x_e)^3$

where;

$x_e$ e = equilibrium position of the end of the spring

i.e. $x_e$ = 0

q = is the spring constant

Also;

$F_x = -qx^3$

From above the units of q can be calculated by making q the subject of the formula :

$q = (F_x)/(x^3)$

where x is in meters and F is in Netwon;

Then :

q = N/m³

The Potential energy P.E of the compressed spring can be calculated by using the integral:

$F_x = -(dU)/(dx)$

$U(x) = - \int F_x dx$

$U(x) = - \int ^x_0 (-qx^3) \ dx$

$\mathbf{U(x) = -(qx^4)/(4)}$

You take a running jump off the end of a diving platform at a speed of 7 m/s and splash into the water 1.5 seconds later.a.) How far horizontally do you land from your takeoff point?
b.) How high is the diving platform?

Answers

Part a

Just use v=d/t

d = vt
d = 7 x 1.5

For part b

Vf=Vi +at
Vf=0 + (9.8)(1.5)

d= ((Vi + Vf) /2) x t
d = ((0 + Vf) /2) x 1.5

In second equation, use Vf found in first equation

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The change in pitch of a train's horn as it passes while you are standing still can be explained by A. the Doppler effect. B. ultrasonic vibrations. C. sympathetic vibration. D. resonance.

Answers

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What is the potential energy of spring?

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The change in pitch of a train's horn as it passes while you are standing still can be explained by A. the Doppler effect.
B. ultrasonic vibrations.
C. sympathetic vibration.
D. resonance.