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exis [7]
3 years ago
12

Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache

d to hanging blocks. The block attached to string 1 has a mass of 20 kg and the block attached to string 2 has a mass of M. Listeners hear a beat frequency of 2 Hz when string 1 is excited at its fundamental frequency and string 2 is excited at its third harmonic. What is one possible value for mass M
Physics
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

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ASHA 777 [7]

Answer:

B) 2g

Explanation:

<u>Given the following data;</u>

Velocity, v = 14m/s

Radius, r = 10m

To find the centripetal acceleration;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Acceleration, a = \frac {14^{2}}{10}

Acceleration, a = \frac {196}{10}

Acceleration, a = 19.6m/s²

In terms of acceleration due to gravity, g = 9.8m/s²

We would divide by g;

Acceleration, a = 19.6/9.8 = 2

Hence, centripetal acceleration = 2g

Therefore, the rider's centripetal acceleration in terms of g, the acceleration due to gravity is 2g.

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3 years ago
An object is suspended by a string from the ceiling of an elevator. If the tension in the string is equal to 25 N at an instant
Phantasy [73]

By Newton's second law, the net force on the object is

∑ <em>F</em> = <em>T</em> - <em>mg</em> = - <em>ma</em>

where

• <em>T</em> = 25 N, the tension in the string

• <em>m</em> is the mass of the object

• <em>g</em> = 9.8 m/s², the acceleration due to gravity

• <em>a</em> = 2.0 m/s², the acceleration of the elevator-object system

Solve for <em>m</em> :

25 N - <em>m</em> (9.8 m/s²) = - <em>m</em> (2.0 m/s²)

==>   <em>m</em> = (25 N) / (9.8 m/s² - 2.0 m/s²) ≈ 3.2 kg

4 0
3 years ago
Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of launch is 75.0° and 15.0° resp
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zaharov [31]
Mass m = 25 Kg;  Force = Weight = mg   Force = ma  but a = g

          F = 40.8 N

Required: gravity g = ?

                F = mg

                g = F/m

                g = 40.8 N/25 Kg

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4 years ago
In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s
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Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

a = \dfrac{1.2 \times 10^5}{86}

a =1395.35\ m/s^2

Using equation of motion

v² = u² + 2 a s

s =\dfrac{v^2}{2a}

s =\dfrac{52^2}{2\times 1395.35}

s = 0.9689 m

The minimum depth of snow that would have stooped him is  s = 0.9689 m

8 0
3 years ago
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