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3 years ago

Will Give Brainliest and 25 Points

Physics

1 answer:

Vikentia [17]3 years ago

5 0

This is the Doppler effect.

1. As the sound leaves the horn the sound waves are at first close to each other and as they move outwards they become further apart. The closer the sound waves are the louder the noise.

As the car gets the closer the sound waves get closer, so the horn becomes louder.

2. As the horn moves away, the sound waves become less frequent, causing the pitch to get lower.

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In order for water to condense on an object, the temperature of the object must be ______ the dewpoint temperature.

enyata [817]

Answer:

at ( or below)

Explanation:

at the dewpoint.......water will condense out of the air onto the surface

7 0

2 years ago

A 17Kg sitting on a shelf has a potential energy of 350J. How high is the shelf? Round your number to the nearest whole number

saul85 [17]

Answer:

${ \bf{potential \: energy = mass \times gravitation \times height}} \\ 350 = 17 \times 10 \times h \\ 350 = 170h \\ height = 2.06 \: metres$

6 0

3 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago

Las carreras de velocidad pura en el atletismo son:

zheka24 [161]

La respuesta es la letra b

8 0

3 years ago

Somebody help I’m stuck on this question <br><br> Work and the work-energy principle

garik1379 [7]

What’s the question?

6 0

3 years ago