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2 years ago

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

Physics

1 answer:

Leokris [45]2 years ago

4 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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If a force of 12 N is applied to an object with a mass of 2 kg, what will be its acceleration?

nadezda [96]

F = ma, Where F is force is in N, mass is in kg, = 2kg, a is acceleration in m/s²

12 = 2a

2a = 12

a = 12/2

a = 6

Acceleration = 6 m/s²

7 0

3 years ago

A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a holl

Semenov [28]

Answer:

a) v=23.9 m/s

b) R=2646 N

Explanation:

According to Newton's second law the net force at the top is given by

∑F=-m*ac

according to

$a_{c}=\frac{v^2}{r}$

∑F= $-m*\frac{v^2}{r}$

a).

To lose contact this means that R=0 so the final equation is

$R-m*g=-m*\frac{v^2}{r}$

$-m*g=-m*\frac{v^2}{r}$

Solve to v

$v^2=g*r$

$v=\sqrt{g*r}=\sqrt{9.8*14.6}$

$v=11.96\frac{m}{s}$

b).

v is the twice of part a so

$v=2*11.96$

$v=23.9\frac{m}{s}$

$R_{m}-m_{m}*g-m_{p}=m*\frac{v^2}{r}$

Solve to Rm

$R_{m}=m*\frac{v^2}{r}+(m_{m}+m_{p})*g$

$R_{m}=m*\frac{23.9^2}{14.6}+(40+70)*9.8$

$R_{m}=2646 N$

5 0

2 years ago

A proton is traveling horizontally to the right at 4.20×10^6m/s.Part A:Find (a)the magnitude and (b) direction of the weakest el

anygoal [31]

Answer:

2630250 N/C, horizontally left

$1.67\times 10^{-8}\ s$

1434.825 N/C, horizontally left

Explanation:

m = Mass of particle

u = Initial velocity = $4.2\times 10^6\ m/s$

v = Final velocity = 0

t = Time taken

s = Displacement = 3.5 cm

q = Charge of particle = $1.6\times 10^{-19}\ C$

Force is given by

$F=qE$

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{1.67\times 10^{-27}}\\\Rightarrow a=95808383.23353E$

$v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 95808383.23353Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 95808383.23353\times 0.035}\\\Rightarrow E=-2630250\ N/C$

Magnitude of electric field is 2630250 N/C

Direction is horizontally to the left

The angle counterclockwise from left is zero.

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times -2630250}{1.67\times 10^{-27}}\\\Rightarrow a=-2.52\times 10^{14}\ m/s^2$

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-4.2\times 10^6}{-2.52\times 10^{14}}\\\Rightarrow t=1.67\times 10^{-8}\ s$

The time taken is $1.67\times 10^{-8}\ s$

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{9.11\times 10^{-31}}\\\Rightarrow a=175631174533.4797E$

$v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 175631174533.4797Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 175631174533.4797\times 0.035}\\\Rightarrow E=-1434.825\ N/C$

Magnitude of electric field is 1434.825 N/C

Direction is horizontally to the left

8 0

2 years ago

The pitch of a sound is most closely related to what ?

Volgvan

It’s related to sound frequency! Hope this helps!

5 0

2 years ago

what net force is required to accelerate a car at a rate of 2 m/s^2 if the car has a mass of 3,000 kg?

zmey [24]

F = ma
F = 2 * 3000
F = 6000N

6 0

3 years ago

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

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