All categories

3 years ago

Which statement describes the reason for the change in the velocity of waves as temperature decreases?

Physics

1 answer:

DIA [1.3K]3 years ago

7 0

Answer:

O The particles of the medium move more slowly and there are fewer chances to transfer energy.

Explanation:

Various media are made up of particles. These particles are in constant motion according to the kinetic theory of matter. Recall that temperature has been defined as the average kinetic energy of the particles in a medium. Hence, for any given medium, the velocity of particle motion increases or decreases linearly with temperature.

The speed of particles in any medium increases or decreases as the temperature of the medium increases or decreases as emphasised above. Hence, at low temperature, the velocity of waves set up by the motion of particles in a medium decreases and transfer the wave energy to neighbouring particles occurs more slowly than at high temperatures.

You might be interested in

Calculate the work done when a 100-W lightbulb is lit for 30 seconds.

Mila [183]

Answer:

3000 J

Explanation:

8 0

3 years ago

What happens to the temperature of a substance while it is changing state?

MArishka [77]

It stays constant, because it's using that energy to change state

3 0

3 years ago

Read 2 more answers

Pls help ASAP. This is astronomy but there isn't an option :)

marin [14]

From largest to smallest they are: Universe, galaxy, solar system, star, planet, moon and asteroid.

Explanation:Let's describe them from smallest to largest. In fact the size order is not exact as there are exceptions.An asteroid is a rocky body which lies in the asteroid belt between Mars and Jupiter. They are typically quite small object. The largest asteroid Ceres has been reclassified as a dwarf planet.A moon is typically a rocky body which is in orbit around a planet. Some moons such as our Moon are quite large and are typically bigger than asteroid. Some moons can actually be smaller than some asteroids.A planet is a nearly spherical body which is in orbit around the Sun. Planets are larger than moons.A star is what planets orbit around. It is the source of light and heat. Our Sun is a star which is many times bigger than all of the planets.A solar system is a star and all of its planets, asteroids, comets and other bodies. It is significantly bigger than a star.A galaxy, such as our Milky Way Galaxy, is a collection of solar systems orbiting around a central core. Most galaxies have a supermassive black hole at their centres.Galaxies also form clusters which are large scale structures.The universe is everything. It contains billions of galaxies. Lots of information RIGHT!!!!

YOUR VERY WelCoMe!!!! :) :) :) :) :0 :)

7 0

3 years ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0

3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

3 years ago