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2 years ago

12

Why is the weight of the ruler not include in the calculation of moment?

1 answer:

Feliz [49]2 years ago

3 0

Answer:

Since the distance between the pivot and its weight is zero, the moment of its weight about the pivot (= weight × 0) is zero. Hence, the weight of the ruler can be ignored.

Hope this helps:)

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On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze

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Ey64udnhsufifjyueuhhekr

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3 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7

Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I= $\frac{P_{avg}}{A}$

I= $\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}$

I= $7.296\times 10^{-6} W/m^2$

Amplitude of electric field at receiver end

$E_{max}=\sqrt{2I\mu _0c}$

Amplitude of induced emf

= $E_{max}d$

= $\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75$

= $17.591\times 10^{-2}=0.1759 v$

7 0

3 years ago

The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

Verizon [17]

Answer:

The value is $E_i = 1.5596 *10^{-18} \ J$

Explanation:

From the question we are told that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$

The velocity is $v = 2.371*10^6 \ m/s$

The mass of electron is $m_e = 9.109*10^{-31} \ kg$

Generally the energy of the incident light is mathematically represented as

$E = \frac{h * c}{\lambda}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

h is the Planck constant with value $h = 6.62607015 * 10^{-34 } J\cdot s$

So

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$

Generally the kinetic energy is mathematically represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$

=> $E_k = 2.56 *0^{-18} \ J$

Generally the ionization energy is mathematically represented as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$

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2 years ago

Consider a simple pendulum with a period

lisabon 2012 [21]

Answer:

1. The length is 8.35m

2. The period on the moon is 14.05 secs

Explanation:

1. Data obtained from the question. This includes the following:

Period (T) = 5.8 secs

Acceleration due to gravity (g) = 9.8 m/s2

Length (L) =...?

The length can be obtained by using the formula given below:

T = 2π√(L/g)

5.8 = 2π√(L/9.8)

Take the square of both side

(5.8)^2 = 4π^2 x L/ 9.8

Cross multiply

4π^2 x L = (5.8)^2 x 9.8

Divide both side by 4π^2

L = (5.8)^2 x 9.8 / 4π^2

L= 8.35 m

2. Data obtained from the question. This includes the following:

Acceleration due to gravity (g) = 1.67 m/s2

Length (L) = 8.35m (the length remains the same)

Period (T) =?

The period can be obtained as follow:

T = 2π√(L/g)

T = 2π√(8.35/1.67)

T = 14.05 secs

Therefore, the period on the moon is 14.05 secs

4 0

2 years ago