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2 years ago

How is the behavior of waves affected by a medium?

Physics

2 answers:

tamaranim1 [39]2 years ago

7 0

It is slowed downnnnnnnnnnnnnnnnnn

Vlad1618 [11]2 years ago

6 0

They are slowed down.

even air is a medium and will slow down waves (slightly) compared to a vacuum/ space

if the wave hits the medium at an angle it can be 'bent' (refracted) like light through a glass prism

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When are car velocity is positive and acceleration is negative, what is happening to the cars motion

DanielleElmas [232]

The motion of the car is negative

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2 years ago

What is the force felt by the 64-kg occupant of the car? Express your answer to two significant figures and include the appropri

Yakvenalex [24]

Answer:

The force is $-1.67\times10^{5}\ N$

Explanation:

Given that,

Mass of car = 64 kg

Suppose, a 1400-kg car that stops from 34 km/h on a distance of 1.7 cm.

We need to calculate the acceleration

Using formula of acceleration

$v^2-u^2=2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the formula

$0^2-(34\times\dfrac{5}{18})^2=2\times a\times 1.7\times10^{-2}$

$a=\dfrac{(34\times\dfrac{5}{18})^2}{2\times1.7\times10^{-2}}$

$a=-2623.45\ m/s²$

We need to calculate the force

Using formula of force

$F=ma$

$F=64\times(-2623.45)$

$F=-1.67\times10^{5}\ N$

Negative sign shows the direction of the force is in the direction opposite to the initial velocity.

Hence, The force is $-1.67\times10^{5}\ N$

5 0

2 years ago

Which of the following energy resources does not produce greenhouse

Inessa [10]

Answer:

Geothermal

Explanation:

Geothermal is obtained from the heat from the ground which makes it friendly

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2 years ago

An automobile traveling 89.0 km/h has tires of 62.0 cm diameter. (a) What is the angular speed of the tires about their axles? (

bekas [8.4K]

Answer:

a) 79.7rad/s

b) -18.7rad/s^2

c) 53m

Explanation:

We will use the MKS system of unit, so:

$v=89.0km/h=89.0\frac{km}{h}*\frac{1000m.h}{3600km.s}=24.7m/s\\\\d=62.0cm=62.0cm*\frac{0.01m}{1cm}=0.62m$

now, The angular speed is given by:

$\omega=\dfrac{v}{\frac{d}{2}}\\\\\\\omega=\frac{24.7m/s}{0.31m}=79.7rad/s$

in order to obtain the angular acceleration we have to apply the following formula:

$(\omega_f)^2=(\omega_o)^2+2\alpha*\theta\\\\\alpha=-\frac{(\omega_o)^2}{2*rev*2\pi}\\\\\alpha=-\frac{(79.7m/s)^2}{2*27*2\pi}=-18.7rad/s^2$

The linear displacement is given by:

$d_l=\theta*r\\d_l=rev*2\pi*\frac{d}{2}\\\\d_l=27*2\pi*0.31m=53m$

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2 years ago

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Nuetrik [128]

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2 years ago