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VikaD [51]
2 years ago
12

the atomic number of cesium (Cs) is 55. If an atom of cesium has 78 neutrons, what is the atomic mass of cesium?

Physics
2 answers:
Slav-nsk [51]2 years ago
7 0
Atomic mass= number of protons + number of neutrons
55 + 78 = 133
hope this helps
Lilit [14]2 years ago
6 0

Answer:

133

Explanation:

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F = 3x10^8m/s / 6x10^-7m = Hz
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Rudik [331]

Answer: fluid fraction

Explanation:

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3 years ago
A crane lifts a 1,750 kg mass using a steel cable whose mass per unit length is 0.88 kg/m. What is the speed of transverse waves
Sauron [17]

Answer:

139.6m/s

Explanation:

Calculate the tension first, T=m*g

mass(m): 1750kg, gravity(g): 9.8m/s^2

T= 1750*9.8

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Then calculate the wave speed using the equation v = √ (T/μ)

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 =139.6m/s

4 0
2 years ago
The acceleration of a block attached to a spring is given by a=−(0.324m/s2)cos([2.50rad/s]t) a = − ( 0.324 m / s 2 ) c o s ( [ 2
allsm [11]

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Looks like you have:

a = -.324 cos 2.5 t

In this case   ω^2 A = .324

ω = 2.5

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5 0
2 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
2 years ago
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