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3 years ago

15

A 0.80 kg object tied to the end of a 2.0 m string swings as a pendulum. At the lowest point of its swing, the object has a kine

tic energy of 10 J. Determine the speed of the object at the instant when the string makes an angle of 50o with the vertical.

1 answer:

devlian [24]3 years ago

6 0

Answer:

3.32 m/s

Explanation:

From the law of conservation of energy, the sum of mechanical and kinetic energy should be equal to the 10 J given. Potential energy is given by mgh where m is mass, g is acceleration due to gravity and h is the height. For this case, $h= l(1-cos\theta)$ and l is string length, given as 2 m, \theta is given as 50 degrees. Kinetic energy is given by $0.5mv^{2}$ and it is this velocity that is unknown.

$10J=0.5\times 0.8kg\times v^{2}+ 0.8kg\times 9.81\times 2m(1-cos 50^{\circ})\\v\approx 3.32 m/s$

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Find the wavelength λ of the 80.0-khz wave emitted by the bat. express your answer in millimeters.

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4.29 millimeters

Explanation:

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A sound wave has a frequency of 500 Hz and a wavelength of 1.8 m. What is the wave speed of the sound wave? Question 1 options:

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Answer:

The wave speed of the sound wave is 900 $\frac{m}{s}$ .

Explanation:

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

The propagation velocity is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement. Relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation: v = f * λ.

In this case:

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One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

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Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$

Solving the above equation for $v_{U-235}$ we have:

$v_{U-235} = 2.68 \cdot 10^{5} m/s$

And by entering that value into equation (1):

$v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s$

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

$E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J$

For U-235:

$E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J$

I hope it helps you!

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3 years ago

How do i do a Wavelength and frequency problem

KengaRu [80]

Answer:

wavelength = v/f or wavelength equals to velocity over frequency

frequency= v/w or velocity over wavelength

frequency= 1/p or one over period or time

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