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A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values

Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or $\lambda=20.0cm$

c. The period, or T, of a wave is found in the equation

$f=\frac{1}{T}$ were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

$50.0=\frac{1}{T}$ so

$T=\frac{1}{50.0}$ and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

$f=\frac{v}{\lambda}$ and since we already have the frequency and we solved for the wavelength already, filling in:

$50.0=\frac{v}{20.0}$ and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0

2 years ago

A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizont

PIT_PIT [208]

Answer:

$v' = 2.83 m/s$

Explanation:

Velocity of wave in stretched string is given by the formula

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

here we know that

T = 4 N

also we know that linear mass density is given as

$\mu = 1 kg/m$

so we have

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

now the tension in the string is double

so the velocity is given as

$v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s$

$v' = 2.83 m/s$

4 0

3 years ago

A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed

Katarina [22]

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object $m_1=52 kg$

distance of $m_1$ from the axle $r_1=14 cm$

mass of rock $m_2=123 gm$

Length of rod $=4.1 m$

distance of $m_2$ from axle $r_2=4.1-0.14=3.96 m$

Net torque acting is

$T_{net}=m_1gr_1-m_2gr_2$

$T_{net}=52\times 0.14\times g-0.123\times 3.96\times g$

$T_{net}=6.793\times 9.8$

$T_{net}=66.57 N-m$

Work done by $T_{net}$ is converted to rock kinetic Energy

thus

$T_{net}\times \theta =\frac{mv^2}{2}$

Where $\theta =angle\ turned =\frac{\pi }{2}$

$v= velocity\ at\ launch$

$66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}$

$v^2=66.57\times \pi$

$v=\sqrt{1700.511}$

$v=41.23 m/s$

3 0

3 years ago

In space movies, spacecrafts explode and oxygen is needed for a fire to start. Why is it wrong? What would really happen?

Triss [41]

Answer:

Hey

The reason these "space movies" are wrong is because objects in "space" (not in a "space" craft) can't catch on fire because there is no air in "space".

6 0

3 years ago

During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the

natita [175]

Answer:

Explanation:

If the work done on the cart is NET work

Then the work will result in an increase in kinetic energy

KE₀ + W = KE₁

½mv₀² + W = ½mv₁²

½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²

v₁ = 1.626991...

v₁ = 1.6 m/s

4 0

2 years ago

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