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IRISSAK [1]
3 years ago
11

A device called oacillatwr is used to swnd waves along a streatched string to send waves along a streatched string. The string i

s 20 cm long, and four complete waves fit along its length when the oscillation vibrates 30 times per second the waves on the string.. What is its frequency and speed and wavelength
Physics
1 answer:
vitfil [10]3 years ago
3 0

Answer:

Frequency = 2m/s1m=2 waves/s, or 2 Hz

A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz.

Explanation:

wd Calculating Wave Frequency or Wavelength from Wave Speed

The equation for wave speed:

Frequency = SpeedWavelength or Wavelength = SpeedFrequency

Therefore, if you know the speed of a wave and either the wavelength or wave frequency, you can calculate the missing value.

For example, suppose that a wave is traveling at a speed of 2 meters per second and has a wavelength of 1 meter. Then the frequency of the wave is:

- Frequency = 2m/s1m=2 waves/s, or 2 Hz

A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz. What is its wavelength?

- Substitute these values into the equation for wavelength:

- Wavelength = 2m/s2waves/s=1 m

See more: https://www.ck12.org/physics/wave-speed/lesson/Wave-Speed-MS-PS/

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Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a
____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, v=2.79\times 10^8\ m/s

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s

Time period of oscillation measured by the observer is :

t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s

So, the time period of oscillation measured by the observer is 5.79 seconds.

3 0
3 years ago
A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The
luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

I =\frac{1}{2}mR^2

For calculating the rotational inertia of the cylinder ; we have;

I = \frac{1}{2}m_pR^2

I = \frac{1}{2}*10.53*(0.96)^2

I=5.265*(0.96)^2

I=4.852224

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = \frac{Ia}{R^2}

a = \frac{g}{1+\frac{I}{mR^2}}

a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}

a = 4.1713 m/s²

Using the equation of motion

v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s

3 0
3 years ago
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An object weighing 4 newtons swings on the end of a string as a simple pendulum. At the bottom of the swing, the tension in the
Anit [1.1K]

Answer:

(B) 0.5 g

Explanation:

Newton's second law says ∑  F i = m a .

the rate of change in momentum of a body is proportional to the force applied on the body.

f∝ma

f=kma

were k is constant and equal to 1

The centripetal acceleration is an acceleration.

the tension on the swing and object weight goes to the left hand side while the centripetal acceleration goes to the right handside

At the bottom of the swing, ΣF = FT – mg = mac;

notice that the tension in the swing is 1.5 times the weight of the object

we can write

1.5mg – mg = mac,

0.5mg = mac

0.5 g=ac

8 0
3 years ago
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PLEASE HELP!! Lab: Electromagnetic Induction
Advocard [28]

I am not sure what this is

5 0
2 years ago
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The center of mass is
PolarNik [594]
D is the best answer. In many physics problems we treat an extended object as if it were a point with the same mass located at the center of mass.
5 0
3 years ago
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