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2 years ago

When a wave strikes and object and bounces off . ( example echo)

Physics

1 answer:

zzz [600]2 years ago

5 0

Answer:

C. reflection

Explanation:

reflection occurs when a wave strikes an object and bounces off of it

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A box slides with uniform acceleration up an incline. The box has an initial speed of 9.0 m/s and rises vertically 2.60 m before

Tju [1.3M]

Answer: 0.58

Explanation:

First we need to get the acceleration of the body using equation of motion

v²=u²-2as

v is the final velocity

u is the initial velocity

a is acceleration

s is the distance moved

0²=9²-2a(2.6)

-81=-5.2a

a=81/5.2

a= 15.6m/s²

Angle of inclination =30°

To get the coefficient of friction, we use the formula

Ff =nR

Ff is frictional force

n is coefficient of friction

R is normal reaction

n = Ff/R = Wsin30°/Wcos30°

n = tan30°

n = 0.58

4 0

3 years ago

When doing numerical calculations involving temperature, you need to pay particular attention to the temperature scale you are u

Mrac [35]

1) $293 ^{\circ}C$

2) $859^{\circ}C$

Explanation:

The average kinetic energy of the molecules of an ideal gas is directly related to the Kelvin temperature of the gas, by the formula

$KE=\frac{3}{2}kT$

where

KE is the kinetic energy

k is the Boltzmann constant

T is the Kelvin temperature

We can say therefore that the average kinetic energy of the particles is directly proportional to the absolute temperature of the gas; so, we can write:

$KE\propto T$

And therefore

$\frac{KE_1}{KE_2}=\frac{T_1}{T_2}$ (1)

In this problem, we have:

$KE_1 = K_{10}$ is the initial kinetic energy of the molecules when the temperature of the gas is

$T_1=10^{\circ}+273=283 K$

Here we want to find the temperature $T_2$ at which the average kinetic energy of the particles is

$KE_2=2K_{10}$

So, twice the initial value. Substituting into eq.(1) and solving for T2, we find:

$T_2=\frac{T_1 KE_2}{KE_1}=\frac{(283)(2K_{10})}{K_{10}}=566 K$

Converting into Celsius degrees,

$T_2=566-273=293 ^{\circ}C$

The root-mean-square (rms) speed of the molecules in a gas is given by the equation

$v=\sqrt{\frac{3kT}{m}}$

where

k is the Boltzmann constant

T is the Kelvin temperature of the gas

m is the mass of each molecule

Therefore, from the equation we can say that the rms speed is proportional to the square root of the temperature:

$v\propto \sqrt{T}$

So we can write:

$\frac{v_1}{v_2}=\frac{\sqrt{T_1}}{\sqrt{T_2}}$ (2)

where in this problem:

$v_1 = v_{rms}$ is the rms speed of the molecules when the temperature is

$T_1=10^{\circ}C+273=283 K$

$v_2=2v_{rms}$ is the final rms speed of the molecules

Solving eq.(2), we find the temperature at which the rms speed is twice the initial value:

$T_2=T_1 (\frac{v_2}{v_1})^2=(283)(\frac{2v_{rms}}{v_{rms}})^2=1132 K$

Converting into Celsius degrees,

$T_2=1132-273=859^{\circ}C$

8 0

3 years ago

Consider a line of children.

defon

Answer:

b,c

Explanation:

Velocity refers to the rate of change of position with respect to time and acceleration refers to the rate of change of velocity with respect to time. Both velocity and acceleration are vector quantities

While the line of children is rotating, <u>the player at the front of the line has the smallest linear velocity and all the children have the same angular acceleration.</u>

7 0

3 years ago

Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v=0.8c

Studentka2010 [4]

Answer:

30.96 m

Explanation:

If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:

d = v * Δt

v = 0.8*c = 0.8 * 3e8 = 2.4e8

Δt = ζ = 129 ns = 1.29e-7 s

d = 2.4e8 * 1.29e-7 = 30.96 m

This is the distance as measured by observer A.

3 0

3 years ago

In a transverse wave, ____________________ is measured from crest to crest or from trough to trough.

PtichkaEL [24]

The blank is wavelength

5 0

3 years ago

Read 2 more answers