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vova2212 [387]
3 years ago
8

How are frequency and period related in simple harmonic motion?.

Physics
1 answer:
xz_007 [3.2K]3 years ago
5 0

They are reciprocals.

That is . . .

-- Frequency  =  1 / period

and

-- Period  =  1 / frequency .

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After a nucleus undergoes radioactive decay, its new mass number is:
Ivanshal [37]
Radioactive "decay" means particles and stuff shoot OUT of a nucleus.
After that happens, there's less stuff in the nucleus than there was before.
So the new mass number is always less than the original mass number.
7 0
4 years ago
Read 2 more answers
A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected
vekshin1

Answer:

(A) 9.14\times 10^{-9}sec

(B) 6.20\times 10^{-3}A

Explanation:

We have given inductance L=5.41\mu H=5.41\times 10^{-6}H

Resistance R=0.949kohm=0.949\times 10^3ohm

Time constant of RL circuit is equal to \tau =\frac{L}{R}

\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

i=i_0(1-e^{\frac{-t}{\tau }})

According to question

0.799i_0=i_0(1-e^{\frac{-t}{\tau }})

e^{\frac{-t}{\tau }}=0.201

{\frac{-t}{\tau }}=ln0.201

{\frac{-t}{5.7\times 10^{-9} }}=-1.6044

t=9.14\times 10^{-9}sec

(b) Current at t=\tau sec

i=i_0(1-e^{\frac{-t}{\tau }})

i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})

i=6.20\times 10^{-3}A

3 0
3 years ago
The sides of a square increase in length at a rate of 3 ​m/sec. a. At what rate is the area of the square changing when the side
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The area of a square is given by:

A = s²

A is the square's area

s is the length of one of the square's sides

Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:

d[A]/dt = d[s²]/dt

The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt

dA/dt = 2s(ds/dt)

A) Given values:

s = 14m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(14)(3)

dA/dt = 84m²/s

B) Given values:

s = 25m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(25)(3)

dA/dt = 150m²/s

6 0
3 years ago
What happens to the speed of the particles if the temperature is increased
dem82 [27]
In a simplistic model, the average speed of the particles increases when temperature is increased.
8 0
4 years ago
An object can be moving from 10 seconds and still have zero displacement true or false
Tasya [4]

It is True :) :( :) :(


8 0
4 years ago
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