The number of electrons in one coulmb charge is ?

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

calculate the period of a wave whose frequency is 5 Hertz and whose wavelength is one centimeter give your answer in a decimal f

olga2289 [7]

The period of the wave is the reciprocal of its frequency.

1 / (5 per second) = 0.2 second .

The wavelength is irrelevant to the period. But since you
gave it to us, we can also calculate the speed of the wave.

Wave speed = (frequency) x (wavelength)

= (5 per second) x (1cm) = 5 cm per second

4 0

4 years ago

Fiziksel büyüklüklerin skaler ve vektörel olarak ayrı ayrı sınıflandırılmasının nedeni nedir?

Nezavi [6.7K]

Uhm sir what language is this

5 0

3 years ago

A system consists of three particles with these masses and velocities: mass 3.00 kg, moving north at 3.00 m/s; mass 4.00 kg, mov

lions [1.4K]

Answer:

the total momentum is 8 .2 kg m/s in north direction.

Explanation:

given,

mass(m₁) 3.00 kg, moving north at v₁ = 3.00 m/s

mass(m₂) 4.00 kg, moving south at v₂ = 3.70 m/s

mass(m₃) 7.00 kg, moving north at v₃ = 2.00 m/s

north as the positive axis

south as the negative axis

now

total momentum = m₁v₁ + m₂ v₂ + m₃ v₃

total momentum = 3 x 3 - 4 x 3.7 + 7 x 2

= 9 - 14.8 + 14

= 8 .2 kg m/s

hence, the total momentum is 8 .2 kg m/s in north direction.

7 0

4 years ago

Calculate the recoil speed of a 1.4 kg rifle shooting 0.006 kg bullets with muzzle speed of 800 m/a.3.43 m/s

Angelina_Jolie [31]

Answer:

a) 3.43 m/s

Explanation:

Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the bullet is shot is zero, because they are both at rest, so:

$p_i = 0$

Instead the total momentum of the system after the shot is:

$p_f = mv+MV$

where:

m = 0.006 kg is the mass of the bullet

M = 1.4 kg is the mass of the rifle

v = 800 m/s is the velocity of the bullet

V is the recoil velocity of the rifle

The total momentum is conserved, therefore we can write:

$p_i = p_f$

Which means:

$0=mv+MV$

Solving for V, we can find the recoil velocity of the rifle:

$V=-\frac{mv}{M}=-\frac{(0.006)(800)}{1.4}=-3.43 m/s$

where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is

a) 3.43 m/s

5 0

3 years ago