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4 years ago

9

A bungee cord with a spring constant of 800 stretches 6 meters at its greatest displacement. How much elastic potential energy d

oes the bungee cord have? The bungee cord has 9600 ⇒ 14,400 J of elastic potential energy.

1 answer:

Vinil7 [7]4 years ago

4 0

Elastic potential energy is given by formula

$U = \frac{1}{2} kx^2$

here we know that

$k = 800 N/m$

$x = 6 m$

Now using above formula we have

$U = \frac{1}{2}(800)(6^2)$

$U = 14400 J$

So elastic potential energy in the chord is 14400 J

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(11) The speed of radio waves is 300 000 000 m/s.

Neko [114]

Answer:

Wavelength = 9.68 meters

Explanation:

Given the following data;

Speed = 300,000,000m/s

Frequency = 31 Megahertz to Hertz = 31 * 10⁶ Hz

To find the wavelength;

Wavelength = speed/frequency

Wavelength = 300,000,000/31,000,000

Wavelength = 9.68 meters

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3 years ago

Compare the strengths of UV light and microwaves (in Hz). Which type of light is more powerful and how do you know?

TEA [102]

Answer:

UV light is more powerful as it has greater energy.

Explanation:

The energy propagated by electromagnetic waves ( light ) through vacuum or medium is known as electromagnetic radiation.

The frequency/wavelength range of electromagnetic radiation is known as electromagnetic spectrum. The electromagnetic spectrum ranging from gamma ray to radio waves.

Frequency range of UV light = ( 8 x 10¹⁴ to 3 x 10¹⁶ ) Hz

Frequency range of Microwaves = ( 300 x 10⁶ to 300 x 10⁹ ) Hz

Ratio of UV light to Microwaves = ( $\frac{8\times10^{14} }{300\times10^{6} }$ to $\frac{3\times10^{16} }{300\times10^{9} }$ )

= ( 2.66 x 10⁶ to 1 x 10⁸ )

Energy of electromagnetic radiation is given by the relation:

E = hν

Here h is plank's constant and ν is frequency.

UV light is more powerful than Microwaves as frequency of UV light is greater than frequency of microwaves. Thus, by the above equation, the energy of UV light is more than energy of Microwaves.

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3 years ago

Two forces that are not equal in size are

Bumek [7]

They are unbalanced forces ..... Hope this helps :3

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3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /

Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

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Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times29.4\times7^2$

$s=720.3\ m$

We need to calculate the velocity

Using formula of velocity

$v=a\times t$

Put the value into the formula

$v=29.4\times7$

$v=205.8\ m/s$

We need to calculate the height

Using formula of height

$H=\dfrac{v^2}{2g}$

Put the value into the formula

$H=\dfrac{(205.8)^2}{2\times9.8}$

$H=2160.9\ m$

We need to calculate the maximum height

Using formula for maximum height

$H'=H+s$

Put the value into the formula

$H'=2160.9+720.3$

$H'=2881.2\ m$

Hence, The maximum height is 2881.2 m.

4 0

3 years ago

La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit

katovenus [111]

Answer:

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Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

v = dr / dt

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r = 2 cos πt i^ + 3 sin πt j^

let us note that it is a movement in two dimensions

let's perform the derivative

v = -2π sin πt i^ + 3π cos πt j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

v = (-4.44 i^ + 6.66 j^ ) m/s

To calculate the mean acceleration we use the expression

a = ( $v_{f} - v_{o}$ ) / Δt

indicates that the time is the first 3 s

we look for the initial velocity t = 0 s

v₀ = 0 i ^ + 3π j ^

we look for the fine velocity, t = 3 s

v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

Δt = (3 -0) = 3 s

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4 years ago