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murzikaleks [220]
3 years ago
6

How many poles do you expect to see in your magnet? Look around your room. Make a list of 5 items that might be attracted to the

magnet: Make a list of 5 items that you think will not interact with the magnet: Check your hypothesis in (1.2) and (1.3) using a magnet. Did the magnet attract all of the items on your list in (1.2)? Did it interact with any of the items in part (1.3)?
Physics
1 answer:
jeyben [28]3 years ago
5 0

Answer:

1.1 Two poles: North and South Poles.

1.2 - Staple pin - Nail - Tip of my phone charger - Metal keys - Cloth Hanger

1.3 - Wooden bed cot - Plastic pen - Game pad - Wooden shelf - Paper - A T-shirt

1.4 Yes

1.5 No

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Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2
GarryVolchara [31]

Answer:

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

Explanation:

Given vibrating system is

u''+\frac{1}{4}u'+2u= 2cos \omega t

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) =  - A ω² cosωt -B ω² sin ωt

Putting this in given equation

-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t

Equating the coefficient of sinωt and cos ωt

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2

\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0.........(1)

and

\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0

\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0........(2)

Solving equation (1) and (2) by cross multiplication method

\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}

\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}   and        B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

5 0
3 years ago
What are the effects of noise?​
slega [8]

Answer:

Noise making has led to loss on hearing.

Explanation:

Supposing you like engaging in parties because of the noise of the sound system it can cause loss on hearing if continued for long

6 0
3 years ago
What is density of a 36g object with a volume of 15
valentinak56 [21]
2.4g

explanation: mass/volume=density
36/15=2.4
7 0
3 years ago
a current of 180 mini amphere passes through a conductor for 5minute calculate the quantity of electricity transported​
oksano4ka [1.4K]

Answer:

Explanation:

You can calculate the total electric charge that passes through the conductor as q=It=(180\times 10^{-3})(5\times 60)= 54 C. It means that the number of electron that passes through the conductor is:

n=\frac{q}{e}=\frac{54}{1.6\times 10^{-19}}=33.75\times 10^{19}

8 0
2 years ago
A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina
Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy U=4.30\times10^{3}\ J

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

\Delta S=\dfrac{\Delta U}{T}

Where, \Delta U = internal energy

T = average temperature

Put the value in to the formula

\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}

\Delta S=-14.72\ J/K

Hence, The approximate change in entropy is -14.72 J/K.

5 0
3 years ago
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