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3 years ago

Si se aumenta la tencion en una cuerda,¿ que pasa con la velocidad de la onda?

Physics

1 answer:

Veronika [31]3 years ago

3 0

Answer:

El aumento de tensión alarga la longitud de onda, reduce la amplitud, aumenta la frecuencia y, por lo tanto, aumenta la velocidad. (GOOGLE)

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Two identical objects (A, B) travel circles of the same radius, but object A completes three times as many rotations as object B

malfutka [58]

Answer:

a) One-ninth the force acting on object A.

Explanation:

First, we derive an expression for the centripetal force acting on both objects.

For object A, centripetal force is:

$F_A = \frac{m{v_A}^2}{r}$

For object B, centripetal force is:

$F_B = \frac{m{v_B}^2}{r}$

We are given that they have the same mass and they move in circles of the same radius.

If object A completes three times as many rotations as object B, then, object must have 3 times the speed of object B.

Hence:

${v_A} = 3*{v_B}$

Therefore, $F_A$ becomes:

$F_A = \frac{m({3*v_B}^{2} )}{r}\\\\\\F_A = \frac{9m{v_B}^{2}}{r}$

$F_A = 9F_B$

=> $F_B = \frac{1}{9} F_A$

Therefore, the net centripetal force acting on object B is one-ninth of the force acting on object A.

4 0

3 years ago

A closed system consists of 0.3 kmol of octane occupying a volume of 5 m3 . Determine

Alenkinab [10]

Answer:

(a). The weight of the system is 336.32 N.

(b). The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

Explanation:

Given that,

Weight of octane = 0.3 kmol

Volume = 5 m³

(a). Molecular mass of octane

$M=114.28\ g/mol$

We need to calculate the mass of octane

Mass of 0.3 k mol of octane is

$M=114.28\times0.3\times1000$

$M=34.284\ kg$

We need to calculate the weight of the system

Using formula of weight

$W=mg$

Put the value into the formula

$W=34.284\times9.81$

$W=336.32\ N$

(b). We need to calculate the molar volume

Using formula of molar volume

$\text{molar volume}=\dfrac{volume}{volume of moles}$

Put the value into the formula

$\text{molar volume}=\dfrac{5}{0.3}$

$\text{molar volume}=16.6\ m^3/k mol$

We need to calculate the mass based volume

Using formula of mass based volume

$\text{mass based volume}=\dfrac{volume}{mass}$

Put the value into the formula

$\text{mass based volume}=\dfrac{5}{34.284}$

$\text{mass based volume}=0.145\ m^3/kg$

Hence, (a). The weight of the system is 336.32 N.

(b). The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

3 0

3 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

3 years ago

A 20.94 g sample of an unknown metal is heated to 99.4 oC in a hot water bath. The metal sample is transferred to a calorimeter

MAXImum [283]

Answer:

1092 J

Explanation:

From the question,

Assuming no heat is lost to the surrounding

Heat lost by the metal = Heat gained by water.

H' = Cm(t₂-t₁)................. Equation 1

Where H' = Heat lost by the metal, C = specific heat capacity of water, m = mass of water, t₂ = Final temperature of the mixture, t₁ = Initial temperature of water

But,

m = D'v................... Equation 2

Where D' = Density of water, v = volume of water.

Given: D' = 1000 kg/m³, v = 100 mL = 100/1000000 = 0.0001 m²

Substitute into equation 2

m = 1000(0.0001)

m = 0.1 kg.

Also given: C = 4200 J/kg.°C, t₁ = 22 °C, 24.6 °C

Substitute into equation 1

H' = 4200×0.1×(24.6-22)

H' = 420(2.6)

H' = 1092 J.

Hence the heat (q) lost by the metal = 1092 J

3 0

3 years ago

The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

Nana76 [90]

Answer:

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

7 0

3 years ago

Si se aumenta la tencion en una cuerda,¿ que pasa con la velocidad de la onda?​

Si se aumenta la tencion en una cuerda,¿ que pasa con la velocidad de la onda?