When would the displacement technique of measuring need to be employed?

A sound wave traveling at 340 m/s is generated by a 480 Hz tuning fork.

Shtirlitz [24]

Answer:

Wavelength = 0.7083 meters

Explanation:

Given the following data;

Speed of wave = 340 m/s

Frequency = 480 Hz

To find how long is the sound wave, we would determine its wavelength;

Mathematically, the wavelength of a waveform is given by the formula;

Wavelength = velocity/frequency

Wavelength = 340/480

Wavelength = 0.7083 meters

8 0

2 years ago

A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b

Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0

2 years ago

How much work was done by a hot air balloon to lift up a 100 Newton to a height of 300 meters?

Monica [59]

So, the work was done by that hot air-balloon is <u>30,000 J or 30 kJ</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

$\boxed{\sf{\bold{W = F \times s}}}$

With the following condition :

W = work (J)
F = force (N)
s = shift or displacement (m)

Now, the s (displacement) can be written as ∆h (altitude change) because the object move to vertical line. The formula can also be changed to:

$\boxed{\sf{\bold{W = F \times \Delta h}}}$

With the following condition :

W = work (J)
F = force (N)
$\sf{\Delta h}$ = change of altitude (m)

If an object has mass, then the object will also be affected by gravity. Always remember that F = m × g. So that :

$\sf{W = F \times \Delta h}$

$\boxed{\sf{\bold{W = m \times g \times \Delta h}}}$

With the following condition :

W = work (J)
m = mass of the object (kg)
g = acceleration of the gravity (m/s²)
$\sf{\Delta h}$ = change of altitude (m)

<h3>Problem Solving</h3>

We know that :

F = force = 100 N
$\sf{\Delta h}$ = change of altitude 300 m

What was asked :

W = work = ... J

Step by step :

$\sf{W = F \times \Delta h}$

$\sf{W = 100 \times 300}$

$\boxed{\sf{W = 30,000 \: J = 30 \: kJ}}$

<h3>Conclusion</h3>

So, the work was done by that hot air-balloon is 30,000 J or 30 kJ.

Work that he had done to lift object brainly.com/question/26341717
Converting work to potential energy brainly.com/question/26487284

5 0

2 years ago

What is the mass of a falling rock if it produces a force of 50 N?

Rasek [7]

×× $F=m \times a$ ×

50N is your force and the acceleration is -9.8m/s^2 due to gravity.

So, you just plug that in.

$50 N=m\times-9.8m/s^2\\
\frac{50}{-9.8}=m\\
m=-5.102$

BUT you know that mass cannot be negative, so you just disregard the negative sign and the mass of the rock is 5.102 grams.

8 0

3 years ago

Imagine another solar system, with a star of the same mass as the Sun. Suppose a planet with a mass twice that of Earth (2 Earth

sashaice [31]

Answer:

Time period, T = 403.78 years

Explanation:

It is given that,

Orbital distance, $a=1\ AU=1.496\times 10^{11}\ m$

Mass of the Earth, $m_e=5.972\times 10^{24}\ kg$

Mass of the planet, $m_p=2m_e=11.944\times 10^{24}\ kg$

Let T is the orbital period of this planet. The Kepler's third law of motion gives the relation between the orbital period and the orbital distance.

$T^2=\dfrac{4\pi^2}{Gm_p}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 11.944\times 10^{24}}\times (1.496\times 10^{11})^3$

$T=1.28\times 10^{10}\ s$

or

T = 403.78 years

So, the orbital period of this planet is 404 years. Hence, this is the required solution.

4 0

2 years ago