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leonid [27]
3 years ago
10

A 0.80 kg mass attached to an ideal spring oscillates horizontally with a period of 0.50 s and amplitude of 0.30 m. What is the

spring constant of the spring?
Physics
1 answer:
sasho [114]3 years ago
8 0

Answer:

126.33 N/m

Explanation:

Angular frquency is given by

f=\frac {2\pi}{t}

Wheret is period of wave and f is angular frequency.

Given period of 0.5 s then

w=\frac {2\pi}{0.5}=4\pi rad/s\approx 12.57 rad/s

Angular frequency is also given by

w=\sqrt{\frac {k}{m}}

Making k the subject then k=mw^{2}

Where k is spring constant, m is mass

Substituting m with 0.8 then

k=0.8\times (4\pi)^{2}

K=126.330936333944 N/m

Rounded off as 126.33 N/m

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An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
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Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

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3 years ago
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Answer:

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First you need to understand the differences between physical and chemical change; physical and chemical property.

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Chemical change, is different by the fact that they change in chemical composition. Bonds are broken and/or made through the reaction, which changes them not only physically but chemically as well. Some of the most indicative signs of a chemical change occurring are: change in color, odor, production of gas, production of light/heat.

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