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diamong [38]
4 years ago
10

5 A \machine produces compression waves in a spring that is 120 cm long by pulsing twice every second. The back and forth moveme

nt of the pulse creates compressed sections of the spring that are 30 cm apart and travel toward the other end of the spring. Which property of the wave does the value of two per second describe? A. wavelength B. intensity C. frequency D. amplitude
Physics
2 answers:
Lisa [10]4 years ago
8 0

Answer:

D. Frequency

Explanation:

The machine produces compression waves in a spring that is 120cm long and it pulse twice every seconds. The property the wave exhibits by exhibiting the value of two per second is the Frequency.

Amplitude of a a wave is the maximum displacement from the equilibrium position. The amplitude of an object is equal to one half the path of the vibration path. The amplitude of a wave is proportional to the amplitude of the source. Amplitude measures how big a wave is . Amplitude of a wave usually look at the energy of a wave so it is not the amplitude .

The frequency of a wave gives the number of  wave that passed fixed point at a given time.

olga2289 [7]4 years ago
7 0
The correct answer is C) frequency.
In fact, the frequency is the number of wave crests (or pulses) per seconds. In our problem, the machine that produces the wave pulses two times per second, so this is exactly the frequency of the compression wave.
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A 18000 Hz wave has wavelength of 0.03 meters. How fast is this wave<br> traveling?
Alex Ar [27]

Answer:

540m/s

Explanation:

Given parameters:

Frequency of the wave = 18000Hz

Wavelength of the wave = 0.03m

Unknown:

How fast is the wave traveling = ?

Solution:

How fast the wave is traveling is a measure of the speed of the wave;

     Speed of wave  = frequency x wavelength

Now insert the given parameters and solve;

   Speed of wave  = 18000 x   0.03  = 540m/s

4 0
3 years ago
A tennis ball is dropped from 1.43 m above the
Rudiy27

Answer:

-5.29 m/s

Explanation:

Given:

y₀ = 1.43 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.43 m)

v = -5.29 m/s

4 0
4 years ago
8.
GuDViN [60]

This ratio (Fnet/m) is sometimes called the gravitational field strength

and is expressed as 9.8 N/kg ⇒ answer D

Explanation:

The gravitational field strength at a point is:

  • The gravitational force exerted per unit mass placed at that point.
  • This means that the gravitational field strength, g is equal to the force experienced by a mass of 1 kg in that gravitational field
  • Gravitational field strength = Weight/mass
  • Its unit is Newton per kilogram
  • Gravitational field strength ≈ 9.8 N/kg

From the notes above

The ratio \frac{F_{net}}{mass} = Gravitational field strength (g)

The answer is:

This ratio (Fnet/m) is sometimes called the gravitational field

strength and is expressed as 9.8 N/kg

Learn more:

You can learn more about gravitational field strength in brainly.com/question/6763771

LearnwithBrainly

5 0
3 years ago
you weight 650 N. What would you wieght if the Earth were four times as massive as it is and its raduis were three times its pre
LiRa [457]

To solve this problem we will apply the concept given by the law of gravitational attraction, which properly defines gravity under the function

g = \frac{GM}{r^2}

Here,

G = Gravitational Universal Constant

M = Mass of Earth

r = Distance between the human and the center of mass of the Earth

The acceleration due to gravity when is 4 times the mass of Earth and 3 times the radius would be given as,

g_1 = \frac{GM}{r^2}

g_2 = \frac{G(4M)}{(3r)^2}

g_2 = \frac{4GM}{9r^2}

g_2 = \frac{4}{9} g_1

The weight is defined as

W = mg_1

So the new weight would be given as

W' = mg_2

W' = m(\frac{4}{9} g_1 )

W' = \frac{4}{9} mg_1

W' = \frac{4}{9} W

W' = \frac{4}{9}(650)

W' = 288.8N

Therefore the weight under this condition is 288.8N

5 0
3 years ago
An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign
saul85 [17]

Answer:

2.1\times 10^{-9} T

Explanation:

We are given that

Frequency,f=800KHz=800\times 10^{3} Hz

1kHz=10^{3} Hz

Distance,d=4.5 km=4.5\times 10^{3} m

1 km=1000 m

Electric field,E=0.63V/m

We have to find the magnetic field amplitude of the signal at that point.

c=3\times 10^8 m/s

We know that

B=\frac{E}{c}

B=\frac{0.63}{3\times 10^8}=0.21\times 10^{-8} T

B=2.1\times 10^{-9} T

8 0
4 years ago
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