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3 years ago

7

The mass of a mass-and-spring system is displaced 10 cm from its equilibrium position and released. A frequency of 4.0 Hz is obs

erved. What frequency would be observed if the mass had been displaced only 5.0 cm and then released?

2 answers:

scoundrel [369]3 years ago

7 0

Answer:

It will remain same i.e. 4.0 Hz

Explanation:

In a mass-and-spring system, the frequency depends upon the mass (m) and the spring constant (k).

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Since, the frequency does not depend on the initial displacement of the mass, the frequency would remain the same i.e. 4.0 Hz.

Tanya [424]3 years ago

7 0

Answer:

f = 4.0 Hz

Explanation:

The frequency of SHM in spring mass system is given as

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

now we can see that the frequency of SHM of spring mass system depends upon following parameters

1) mass of the object

2) spring constant

So here it does not depends on the displacement of the object from its mean position

So here the frequency will remain the same

So here

f = 4.0 Hz

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A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$

The normal force N is equal to the weight of the car, thus

$F_r=\mu .m.g$

Equating both forces

$\displaystyle \mu .m.g=m.\frac{v^2}{r}$

Simplifying

$\displaystyle \mu =\frac{v^2}{rg}$

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3 years ago

a gym consists of a rectangular region with a semi-circle on each end. if the perimeter of the room is to be a 200 m running tra

Nikolay [14]

The dimensions of the rectangle are:

l = 50 m

b = $100/\pi$ m

<h3>What is a perimeter in math?</h3>

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.

<h3>How do we find a perimeter of a rectangle?</h3>

The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.

<h3>Given:</h3>

As per the question:

Perimeter of the room is given as P = 200 m

The region is rectangular having a semicircle at each end.

Now,

Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.

Then, Area of the given rectangle, A = lb

Perimeter of the room, P is = $\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l$

Therefore, $\pi b+2l=200$

$b=(200-2l)/\pi$

Now,

Area, A = $l(200-2l)/\pi=(200l-2l^{2} )/\pi$

Now, differentiate A w.r.t l:

Again differentiating w.r.t 'l', we get:

$d^{2} A/dl^{2} =-4l/\pi$ < 0

Thus we get maximum are when $dA/dl=0$

Therefore,

$(200-4l)/\pi=0$

l = 50 m

Now, from

$\pi b+2l=200$

$\pi b=200-2*50$

$b=100/\pi$

$r=b/2=50/\pi$

To know more about area of a recatangle, visit the link

brainly.com/question/20693059

#SPJ4

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