I need help with this question, could you help me? :))) ASAP

A wave pulse travels along a string at a speed of 200 cm/s. What will be the speed if:

34kurt

Answer:

a) $v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s$

The velocity increase by a factor of $\sqrt{2}$

b) $v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s$

The velocity decrease by a factor of 2.

c) $v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s$

The velocity increase by a factor of 2

d) $v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s$

The velocity not changes.

Explanation:

For this case we know that the velocity is $v = 200 cm/s = 2m/s$

$v_f$ represent the final velocity after the changes specified,

Part a

The formula for the speed of a wave in a string is given by:

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

And the linear density is defined as:

$\rho = \frac{m}{L}$

And if we replace this we got:

$v = \sqrt{\frac{TL}{m}}$

If the tension mass is doubled we have this:

$v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s$

The velocity increase by a factor of $\sqrt{2}$

Part b

If we mass is quadrupled we have this:

$v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s$

The velocity decrease by a factor of 2.

Part c

If the length is quadrupled we have this:

$v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s$

The velocity increase by a factor of 2

Part d

For this case we know that the mass and the length are both quadrupled and we got:

$v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s$

The velocity not changes.

7 0

3 years ago

In a process called pair production, an energetic

vredina [299]

Answer:

(3) mass-energy must be conserved

Explanation:

As we know that gamma rays are mass less and charge less photons which will have sufficient energy.

Now in case of Pair production the gamma photons convert its whole energy into mass by the law of Einstein's mass energy equivalence relation.

As per his relation we can say

$E = \Delta m c^2$

here we will have

[tex]\Delta m[\tex] = mass produced

now we also have to think that here the two particles must have a pair of particles and antiparticles so that the combined mass system will have energy equivalent to the energy of gamma photons and also it must follow the conservation of charge

So here it will form an electron and a positron such that total charge will be zero and the energy will be same as energy of gamma photon.

3 0

4 years ago

Read 2 more answers

Janice is unsure about her future career path. She has grown up on her family farm, but she is also interested in medicine. Jani

Vika [28.1K]

Answer:

d not joining FRA and joining HOSA INSTEAD

3 0

4 years ago

Read 2 more answers

What are two limitations of using a marble to model an atom?

Kryger [21]

Answer:

Explanation:

Its c bcyea

4 0

3 years ago

A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a

dem82 [27]

Answer:

16.17 m/s

Explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

$H=\frac{u^{2}Sin^{2}\theta }{2g}$

$H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m$

The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

$v_{y}^{2}=u_{y}^{2}+2gh'$

here, uy = 0

So,

$v_{y}^{2}=2\times 9.8 \times 6.32$

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{11.87^{2}+11.13^{2}}$

v = 16.27 m/s

7 0

4 years ago