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attashe74 [19]
3 years ago
8

using dimensional analysis find the relation between the velocities of transverse waves produced from the vibration of thin homo

geneous string and between the tension in the string and mass per unit length of it.​
Physics
1 answer:
gulaghasi [49]3 years ago
5 0

Answer:

v^2=\frac{T}{M}

where v = transverse wave velocity, T = tension in the string, M = mass per unit length.

Explanation:

Dimensional analysis is where you just look at the units and see how they fit within each other.

In this case, all relationships are made using the MLT comparisons, where M stands for Mass, L stands for Length, and T stands for time.

For example, for velocity, we know the SI unit for velocity is [ms⁻¹] which is L¹T⁻¹, we can do the same thing for tension = [N = Kgms⁻²] = M¹L¹S⁻², and for the mass per unit length which we can think of as [Kgm⁻¹] = M¹L⁻¹.

If you play around a little with the powers, you can find a relationship:

v^2=\frac{T}{M}

since:

[L¹T⁻¹]² = L²T⁻² = M¹L¹T⁻² ÷ M¹L⁻¹ = M¹⁻¹L¹⁻⁽⁻¹⁾T⁻² = M⁰L²T⁻² = L²T⁻²

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B: an increase in acceleration caused an increase in force.

This is based on the concept of force on an object.

Now, formula for force is commonly known as;

Force = mass × acceleration

Now, mass and acceleration are the input values that make the output which is the Force to either increase or decrease.

  • Now, for the line of dominoes to fall, it means that the force was so overwhelming that the dominoes couldn't resist it.

Now, this output which is the force became so much as a result of the increase in acceleration of the motion that triggered its movement.

Moreover, the force is directly proportional to the acceleration!

Thus, we can say that an increase in acceleration caused an increase in force.

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3 years ago
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A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the h
dybincka [34]

Given,

A player kicks a soccer hits at an angle of 30° at a speed of 26 m/s

We can resolute the trajectory of soccer into horizontal and vertical components.(Please see the attached file)

We can have,

Horizontal velocity component of ball= 26cos(30°)  = 26×(√3÷2) = 22.51 m/s

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6 0
3 years ago
The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas
Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 m,/sec^2

(c) Escape velocity on earth is 3.563\times 10^4m/sec

Explanation:

We have given radius of mars R_{mars}=6.9\times 10^3km=6.9\times 10^6m and radius of earth R_{E}=1.3\times 10^4km=1.3\times 10^7m

Mass of earth M_E=5.972\times 10^{24}kg

So mass of mars M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg

Volume of mars V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3

So density of mars d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3

Volume of earth  V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3

So density of earth d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3

(A) So the ratio of mean density \frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735

(B) Value of g on mars

g is given by g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2

(c) Escape velocity is given by

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec

5 0
4 years ago
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A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

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We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

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3 years ago
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Answer:

523269.9\ \text{N/m}

Explanation:

q = Charge

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q_1=25\ \text{C}

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r_2=850\ \text{m}

The electric field is given by

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The electric field at the aircraft is 523269.9\ \text{N/m}

4 0
3 years ago
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