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3 years ago

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3. Si usted duplicara la amplitud de un M.A.S. ¿cómo cambiaría la frecuencia, velocidad máxima, la aceleración máxima y la energ

ía total mecánica?

1 answer:

WITCHER [35]3 years ago

8 0

Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía

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A charge of +2 C is at the origin. When charge Q is placed at 2 m along the positive x axis, the electric field at 2 m along the

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Answer:

The value of Q is - 8 C

Explanation:

Given;

the magnitude of first charge = 2 C

position of the first charge = 0

_ (2m) 0(+2C) +(2m)(Q)

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E₁ --------------------------------->

<------------------------------------------------------------------ E₂

E₁ and E₂ are equal in magnitude but opposite in direction

| E₂ | = | E₁ |

$\frac{K*Q}{r^2} = \frac{K*2}{r^2} \\\\\frac{Q}{(4)^2} = \frac{2}{(2)^2}\\\\\frac{Q}{16} =\frac{2}{4}\\\\\frac{Q}{16} =\frac{1}{2}\\\\2Q = 16\\\\Q = 8 \ C$

Thus, since E₂ is opposite in direction to E₁, the Q = - 8 C

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4 years ago

Two students, Nora and Allison, are both the same age when Allison hops aboard a flying saucer and blasts off to achieve a cruis

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Answer:

23.38 years

Explanation:

Speed of Allison = 0.83 c = v

Time passed according to Allison = Δt = 29.5 years

Time dilation

$\Delta t'=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{29.5}{\sqrt{1-\frac{0.83^2c^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{29.5}{\sqrt{1-0.83^2}}\\\Rightarrow \Delta t'=52.88\ years$

For Nora 52.88 years would have passed

So, their age difference would be 52.88-29.5 = 23.38 years

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4 years ago

Light with a wavelength of 494 nm in vacuo travels from vacuum to water. Find the wavelength of the light inside the water in nm

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Answer:

370.6 nm

Explanation:

wavelength in vacuum = 494 nm

refractive index of water with respect to air = 1.333

Let the wavelength of light in water is λ.

The frequency of the light remains same but the speed and the wavelength is changed as the light passes from one medium to another.

By using the definition of refractive index

$n = \frac{wavelength in air}{wavelength in water}$

where, n be the refractive index of water with respect to air

By substituting the values, we get

$1.333 = \frac{494}{\lambda }$

λ = 370.6 nm

Thus, the wavelength of light in water is 370.6 nm.

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3 years ago

The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from

anygoal [31]

Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) $V_{max}$ the car has is:

$V_{max}=203 mph=90.74 m/s$ taking into account $1 mile=1609.34 m$

And the car's average acceleration $a_{ave}$ is:

$a_{ave}=0.091 g=2.93 ft/s^{2}=0.89 m/s^{2}$

Since:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}$ (1)

Where:

$V_{f}=V_{max}=90.74 m/s$ is the car's final speed (top speed)

$V_{o}=0 m/s$ because it starts from rest

$\Delta t$ is the time it takes to reach the top speed

Finding this time:

$\Delta t=\frac{V_{f}-V_{o}}{a_{ave}}$ (2)

$\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}}$ (3)

$\Delta t=t_{1}=101.95 s$ (4)

Now we have to find the distance $d$ the car traveled at this maximum speed with the following equation:

$V_{f}^{2}=V_{o}^{2} + 2a_{ave} d$ (5)

Isolating $d$ :

$d=\frac{V_{f}^{2}}{2a_{ave}}$ (6)

$d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})}$ (7)

$d=4625.70 m$ (8)

On the other hand, we know the total distance $D$ traveled by the car is:

$D=12.42 miles = 19988.052 m$

Hence the remaining distance is:

$d_{remain}=D-d=19988.052 m - 4625.70 m$ (9)

$d_{remain}=15362.35 m$ (10)

So, we can calculate the time $t_{2}$ it took to this car to travel this remaining distance $d_{remain}$ at its top speed $V_{max}$ , with the following equation:

$V_{max}=\frac{d_{remain}}{t_{2}}$ (11)

Isolating $t_{2}$ :

$t_{2}=\frac{d_{remain}}{V_{max}}$ (12)

$t_{2}=\frac{15362.35 m}{90.74 m/s}$ (13)

$t_{2}=169.45 s$ (14)

With this time $t_{2}$ and the value of $t_{1}$ calculated in (4) we can finally calculate the total time $t_{TOTAL}$ :

$t_{TOTAL}=t_{1}+ t_{2}$ (15)

$t_{TOTAL}=101.95 s + 169.45 s$ (16)

$t_{TOTAL}=271.4 s s$

5 0

4 years ago

Alana is skateboarding at 19 km/h and throws a tennis ball at 11 km/h to her friend Oliver who is behind her leaning against a w

NNADVOKAT [17]

The speed of the tennis ball is 8 km/h when Alana throws the ball behind at the speed of 11 km/h.

<h3>What is the frame of reference?</h3>

It is a point from where the motion is observed. The speed changes with the frame of reference.

For the person standstill on the sidewalk, If Alana throw the ball in the front direction, The speed will be,

19+11 = 30 km/h

It Alana just leave the ball, the speed,

19 +0 = 19 km/h

If Alana throws the ball behind her at the speed of 19 km/h. The speed of the ball for a sidewalk observer,

19 +(-19 ) = 0 km/h

Thus, when Aana throw the ball behind the velocity will be negative,

So,

19 + (-11 ) = 8 km/h

Therefore, the speed of the tennis ball is 8 km/h when Alana throws the ball behind at the speed of 11 km/h.

Learn more about Frame of reference:

brainly.com/question/9820962

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2 years ago