1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tatiana [17]
3 years ago
5

What are (a) the lowest frequency, (b) the second lowest frequency, (c) the third lowest frequency of transverse vibrations on a

wire that is 10.0m long, has a mass of 100g, and is stretched under tension of 250 N?
Physics
1 answer:
Alekssandra [29.7K]3 years ago
8 0

Answer:f1 = 7.90Hz, f2= 15.811Hz, f3 = 23.71Hz.

Explanation: length of string = 10m, mass of string = 100g = 0.1kg, T= 250N

We need the velocity of sound wave in a string when plucked with a tension T, this is given below as

v = √T/u

Where u = mass /length = 0.1/ 10 = 0.01kg/m

Hence v = √250/0.01, v = √25,000 = 158.11

a) at the lowest frequency.

At the lowest frequency, the length of string is related to the wavelength with the formulae below

L = λ/2, λ= 2L.

λ = 2 * 10

λ = 20m.

But v = fλ where v = 158.11m/s and λ= 20m

f = v/ λ

f = 158.11/ 20

f = 7.90Hz.

b) at the first frequency.

The length of string and wavelength for this case is

L = λ.

Hence λ = 10m

v = 158.11m/s

v= fλ

f = v/λ

f = 158.11/10

f = 15.811Hz

c) at third frequency

The length of string is related to the wavelength of sound with the formulae below

L =3λ/2, hence λ = 2L /3

λ = 2 * 10 / 3

λ = 20/3

λ= 6.67m

v = fλ where v = 158.11m/s, λ= 6.67m

f = v/λ

f = 158.11/6.67

f = 23.71Hz.

You might be interested in
Which of the following has to happen for Earth's shadow to fall on the moon?
Ronch [10]
The correct answer is a, lunar eclipse
6 0
2 years ago
Why do astronauts' weights differ in space?
OLga [1]

Answer:

There is less gravitational force in space.

Explanation:

Gravity doesn't exist as it does on Earth. Earth's gravity is as 6 times stronger as it is on the moon.

7 0
3 years ago
Two black holes (the remains of exploded stars), separated by a distance of
jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

  • <em>distance between the two black holes, r = 10 AU = 1.5 x 10¹² m</em>
  • <em>gravitational force between the two black holes, F = 6.9 x 10²⁵ N.</em>
  • <em>combined mass of the two black holes = 5.20 x 10³⁰ kg</em>

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁  + 2.328 x 10⁶⁰ = 0

<em>solve the quadratic equation using formula method</em>;

a = 1, b =-  5.2 x 10³⁰, c = 2.328 x 10⁶⁰

m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20})  \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg  -  4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

or

m₂ = 5.2 x 10³⁰ kg  -  4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

Learn more here:brainly.com/question/9373839

3 0
2 years ago
A steady 45 N horizontal force is applied to a 15 kg object a table. The object slides against a friction force of 30 N. Calcula
lisov135 [29]

The net force on an object subject to friction is equal to the sum of the applied force and the frictional force.

Mathematically,

F_{N} = ma = F_{applied} - f_{fr}

Here, m is mass of object and a is its acceleration. We take frictional force negative because it opposes the motion of object.

Given, m=15\ kg , F_{applied} =45\ N and f_{fr} = 30\ N

Substituting these values in above formula, we get

15\ kg\times a = 45\ N -30\ N=15\ N \\\\a=\frac{15\ N}{15\ kg} =1\ m/s^2.

Thus, the acceleration of an object is 1\ m/s^2.


6 0
3 years ago
Một chiếc thuyền có tốc độ 15 km / h trong nước tĩnh lặng đi 30 km về phía hạ lưu và trở lại trong tổng cộng 4 giờ 30 phút. Tốc
Ad libitum [116K]
I don't know this answer
3 0
3 years ago
Other questions:
  • Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
    6·1 answer
  • A rock with a mass of 8 kg falls straight down from a height of 7 m. What work is done?
    8·1 answer
  • What is absolute zero? What is the temperature of absolute zero on the Kelvin and Celsius scales?
    9·2 answers
  • Which of the following could be the source of resistance in a household electric circuit?
    10·2 answers
  • What distinguishes a tornado watch from a tornado warning?
    15·2 answers
  • How does a dry cell produce electricity​
    12·1 answer
  • There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________
    14·1 answer
  • When is the object at rest? in constant speed? accelerating?​
    9·1 answer
  • WHO IS GOOD AT PHYSICS, ELECTRIC CIRCUITS?​
    6·1 answer
  • Which is heavier a pound of bricks or a pound of feather
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!