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2 years ago

A spring vibrates 120 times in 2 mins find its frequency and time period

Physics

1 answer:

german2 years ago

3 0

2 minutes is 120 seconds, so if you were finding vibrations per minute, it would be 60 times a minute.

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A 97 kg man lying on a surface of negligible friction shoves a 62 g stone away from himself, giving it a speed of 2.6 m/s. What

tangare [24]

Answer:

man will move in opposite direction with speed

$v_1 = 1.66 \times 10^{-3} m/s$

Explanation:

As we know that man is lying on the friction-less surface

so here net force along the surface is zero

so if we take man + stone as a system then net change in momentum of this system will become zero

so here we have

$P_i = P_f$

$0 = m_1v_1 + m_2v_2$

here we have

$0 = (97)v_1 + 0.062(2.6)$

$v_1 = -\frac{0.1612}{97}$

$v_1 = -1.66 \times 10^{-3} m/s$

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2 years ago

How do u answer this?

gayaneshka [121]

Answer:

food

Explanation:

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4 0

3 years ago

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis

klio [65]

Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

$d = v_0t + \frac{1}{2}at^2$

The block starts from rest, so:

$d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2$

Now, we can do a summation of forces of the block using Newton's Second Law:

$F = ma = m_bg - T$

mb = mass of the block

T = tension of string

Solve for tension:

$T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N$

Now, we can do a summation of torques for the wheel:

$\Sigma \tau = rF\\\\\Sigma\tau = rT$

Rewrite:

$I\alpha = rT$

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

$\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2$

Now, plug in the values into the equation:

$I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}$

8 0

2 years ago

A flea walking along a ruler moves from the 45 cm mark to the 27 cm mark. It does this in 3 seconds. What is the speed? What is

Alekssandra [29.7K]

Answer:

Speed= 6cm/s and velocity= 6cm/s in the negative direction

Explanation:

the change in position is from 45cm to 27 cm (moving towards the negative x direction)

$\Delta x = 45 cm - 27 cm = 18 cm$

And the change in time:

$\Delta t= 3 s$

Now we must define the difference between speed and velocity:

Speed is a scalar quantity, which means that it is a number. Velocity is also a number but you must also indicate the direction of the movement.

Thus, the speed is:

$speed= \Delta x/ \Delta t = 18cm/3s=6cm/s$

An the velocity is:

6cm/s in the negative direction

8 0

3 years ago

What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?

PolarNik [594]

Answer:

$\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m$

$\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$n_i=n\ and\ n_f=n-1$

Thus solving it, we get:

$\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m$

So,

$\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m$

Also, $\Delta E=h\times \nu$

So,

$h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}$

$\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

$\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

8 0

3 years ago

A spring vibrates 120 times in 2 mins find its frequency and time period​

A spring vibrates 120 times in 2 mins find its frequency and time period