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worty [1.4K]
3 years ago
15

5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be

changed to raise the wave speed to 180 m/s?
Physics
1 answer:
ANTONII [103]3 years ago
3 0

Answer:

The tension on string when the speed was raised is 134.53 N

Explanation:

Given;

Tension on the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The speed of the wave is given as;

v = \sqrt{\frac{T}{\mu} }

where;

μ is mass per unit length

v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2}  = \frac{T_2}{v_2^2}

The final tension T₂ will be calculated as;

T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N

Therefore, the tension on string when the speed was raised is 134.53 N

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LenKa [72]

Answer:

7.46 J/kg/K

Explanation:

The heat absorbed or lost is:

q = mCΔT

where m is the mass, C is the heat capacity, and ΔT is the change in temperature.

Given q = 15.0 J, m = 0.201 kg, and ΔT = 10.0 °C:

15.0 J = (0.201 kg) C (10.0 °C)

C = 7.46 J/kg/°C

Which is the same as 7.46 J/kg/K.

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3 years ago
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Find the magnitude and direction of a force between a 25.0 coulomb charge and a 40.0coulomb charge when they are separated by a
deff fn [24]

Answer:

95.0 colomb

Explanation:

Make sure to understand the concept

6 0
3 years ago
When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?
Oduvanchick [21]
First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
a = r \times ( \frac{1}{n1}  -  \frac{1}{n2} )
We insert the values

a = 2.18 \times  {10}^{ - 18}  \times ( \frac{1}{ {1}^{2} } -  \frac{1}{ {6}^{2} }  )
= 2.12 \times {10}^{ - 18}
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
wavelength =  \frac{h \times c}{a}
Finally we insert the values
\frac{6.626 \times  {10}^{ - 34}  \times 3 \times  {10}^{8} }{2.12 \times  {10}^{ - 18} }  = 9.376 \times  {10}^{ - 8}
Which is the same as 93.8 nm
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3 years ago
How does increasing the distance between charged objects affect the electric force between them? the electric force increases be
konstantin123 [22]

the electric force decreases because the distance has an indirect relationship to the force

Explanation:

The electric force between two objects is given by

F=k \frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the distance between the two objects

As we can see from the formula, the magnitude of the force is inversely proportional to the square of the distance: so, when the distance between the object increases, the magnitude of the force decreases.

3 0
3 years ago
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Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin
Vedmedyk [2.9K]

Answer:

The correct answer is 2.8*10^{-5}ms^{-1}

Explanation:

The formula for the electron drift speed is given as follows,

u=I/nAq

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is 8.93/63.5 molcm^{-3}. Converting this number to m³ using very elementary unit conversion we get 140384molm^{-3}. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}

if we convert the area from mm³ to m³ we get A=80*10^{-6}m^{2}.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}

which is our final answer.

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