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mafiozo [28]
2 years ago
6

In a lab experiment a light flexible string is wrapped around a solid cylinder with mass M = 9.50 kg and radius R. The cylinder

rotates with negligible friction about a stationary horizontal axis. A mass m = 3.00kg is tied at the free end of the string and release and the mass moves downward. Find the magnitude of the acceleration of the object of mass m.

Physics
1 answer:
Lunna [17]2 years ago
5 0

Answer:3.79 m/s^2

Explanation:

Given

Mass M=9.5 kg

m=3 kg

Net Force is equivalent to \sum F=ma

with tension T in the string    

For mass m

mg-T=ma

T=mg-ma--------1

For cylinder

T\cdot R=I\times \alpha

I for solid cylinder is \frac{2}{5}MR^2 , and \alpha =\frac{a}{R}

thus T=\frac{Ma}{2}----2

Substitute the value of T we get

\frac{Ma}{2}=mg-ma

a(\frac{M}{2}+m)=mg

a=\frac{mg}{\frac{M}{2}+m}

a=3.79 m/s^2

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<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

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b=(200-2l)/\pi

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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

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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

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An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
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1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

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v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

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v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

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