Please help will give brainliest, i might have more later

Which of the following is an example of the Doppler effect? A water bug on the surface of a pond is producing small ripples in t

noname [10]

Answer:

A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.

Explanation:

In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.

Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.

The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.

The phenomenon of Doppler effects is generally applicable to both sound and light.

An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.

Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc.

3 0

3 years ago

A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t

ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0

3 years ago

Paraffin wax is used to make candles true or false

tangare [24]

Answer:

true

Explanation:

PLS MAKE ME AS BRAINLIST

7 0

3 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

3 years ago

g If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of th

Reil [10]

Answer:

Speed; v = 17 m/s

Explanation:

We are given;

Radius; r = 110m

Angle; θ = 15°

Now, we know that in circular motion,

v² = rg•tanθ

Thus,

v = √(rg•tanθ)

Where,

v is velocity

r is radius

g is acceleration due to gravity

θ is the angle

Thus,

v = √(rg•tanθ) = √(110 x 9.8•tan15)

v = √(288.85)

v = 17 m/s

8 0

3 years ago