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3 years ago

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If you are at a train station and a train comes toward you, the train blows its whistle. as the train comes closer the whistle g

ets louder and as the train gets farther the whistle is quieter. What is this effect called?

1 answer:

Wittaler [7]3 years ago

4 0

This is called the Doppler Effect.

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HELP PLEASE!!! 30+ points!!

GarryVolchara [31]

1) 9.26 cm

Explanation:

The focal length of a plane mirror is virtually infinite. Considering the lens equation,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where f is the focal length, p is the object distance, q the image distance. If we replace f with infinity, we get

$q=-p$

The magnification equation states that

$y' = -\frac{q}{p}y$

where y is the size of the object and y' the size of the image. Substituting q=-p, we get

$y'=y$

this means that the image produced by a plane mirror is always:

- Upright (y' is positive)

- The same size as the object

In this case, we have a book of height 9.26 cm (y=9.26 cm). This means that the magnitude of the size of the image (y') will be 9.26 cm as well.

2) 22.7 cm

As we said before, due to the infinite focal length of a plane mirror,

$q=-p$

this means that the image produced by a plane mirror is always:

- Virtual (because q is negative)

- At the same distance from the mirror as the object

In this case, we have a book placed at 22.7 cm from the mirror (p=22.7 cm). This means that the magnitude of the distance of the image from the mirror (q) will be 22.7 cm as well.

3) 1.60 m/s

We said previously that the image produced by a plane mirror is always at the same distance from the mirror as the real object. This implies that whenever we move the object toward/away from the mirror, the distance p will alway remain equal to the distance q. But this also means that the object and the distance are moving toward/away from the mirror at the same speed.

Therefore, since in this case the person is moving away from the mirror at 1.60 m/s, the image will also move away at a speed of 1.60 m/s.

4 0

3 years ago

A bicycle rider has a speed of 19.0 m/s at a height of 55.0 m above sea level when he begins coasting down hill. The mass of the

lukranit [14]

Answer:

The mechanical energy of the rider at any height will be 6.34 × 10⁴ J.

Explanation:

Hi there!

The mechanical energy of the rider is calculated as the sum of the gravitational potential energy plus the kinetic energy. Since there are no dissipative forces (like friction), the mechanical energy of the rider at a height of 55.0 m above the sea level will be the same at a height of 25.0 m (or at any height), because the loss in potential energy will be compensated by a gain in kinetic energy, according to the law of conservation of energy.

Then, calculating the potential and kinetic energy at 55.0 m and 19 m/s, we can obtain the mechanical energy that will be constant:

Mechanical energy = PE + KE

Where:

PE = potential energy.

KE = kinetic energy.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity.

h = height.

Then, the potential energy of the rider will be:

PE = 88.0 kg · 9.81 m/s² · 55.0 m = 4.75 × 10⁴ J

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where "m" is the mass of the object and "v" its velocity. Then:

KE = 1/2 · 88.0 kg · (19.0 m/s)²

KE = 1.59 × 10⁴ J

The mechanical energy of the rider will be:

Mechanical energy = PE + KE = 4.75 × 10⁴ J + 1.59 × 10⁴ J = 6.34 × 10⁴ J

This mechanical energy is constant because when the rider coast down the hill, its potential energy is being converted into kinetic energy, so that the sum of potential energy plus kinetic energy remains constant.

5 0

3 years ago

Need help ASAP!! Thank you..

Nonamiya [84]

1.potential energy 2.kinetic energy 3.electrical energy 4.electrical energy

5 0

3 years ago

Read 2 more answers

The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif

Komok [63]

Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases

The relation between potential difference and the electric field is given by ΔV = E.d

Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.

The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.

The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.

Refer to more about the potential difference here

brainly.com/question/12198573

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6 0

1 year ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago