What are transverse waves and longitudinal waves have? how are they similar and different?

When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these

Zigmanuir [339]

Answer:

a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

v = $\sqrt{ \frac{T}{ \mu } }$

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

m = 2m₀

let's find the new linear density

μ = m / l

iinitial density

μ₀ = m₀ / l

final density

μ = 2m₀ / lo

μ = 2 μ₀

we substitute in the equation for the velocity

initial v₀ = $\sqrt{ \frac{T_o}{ \mu_o} }$

with the new dough

v = $\sqrt{ \frac{T_o}{ 2 \mu_o} }$

v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }

v = 1 /√2 v₀

v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

μ = μ₀

in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

μ = 3m₀ / l₀

μ = 3 m₀/l₀

μ = 3 μ₀

v = $\sqrt{ \frac{T_o}{ 3 \mu_o} }$

v = 1 /√3 v₀

v = 0.577 v₀

d) l = 2l₀

μ = m₀ / 2l₀

μ = μ₀/ 2

v = $\sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }$

v = √2 v₀

v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

μ = 10 m₀/2l₀

μ = 5 μ₀

we look for speed

v = $\sqrt{ \frac{T_o}{5 \mu_o} }$

v = 1 /√5 v₀

v = 0.447 v₀

5 0

3 years ago

Which of the following equations describes Newton’s second law

svp [43]

C.

Newton’s Second Law is F=ma (force is equal to the mass multiplied by acceleration), however, the equation can be rearranged to isolate and calculate mass from force over acceleration. Therefore, m=F/a

8 0

3 years ago

Read 2 more answers

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

So

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

Why is it important to select a coordinate system when studying motion?

kvv77 [185]

Motion is detected when an object changes its position with respect to a reference point. Coordinate system is basically used to represent motion. A coordinate system uses numbers or coordinates which represent position of the reference points on a two-dimensional or three-dimensional space. The trajectory of a point or line can be studied on a coordinate system which describes various aspects of motion like velocity, acceleration, distance, displacement etc. Coordinate system is important because it helps to choose a starting point and the direction (which will be positive).

5 0

3 years ago

A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p

Andrei [34K]

Answer:

13.4 x 10 raise to power -19 C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is E = 214 N/M

, The decrease in electrical potential energy is d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C

5 0

3 years ago