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Juli2301 [7.4K]
3 years ago
15

A clock is designed that uses a mass on the end of a spring as a timing mechanism. If the oscillation time needed is exactly one

second, what is the spring constant required if the mass is 1.3 kg
Physics
1 answer:
user100 [1]3 years ago
5 0

Answer:

The value is  k  =  51.34 \  N/ m

Explanation:

From the question we are told that

   The  mass is  m =  1.3 \  kg

   The needed oscillation time is  T = 1 \ s  

 Generally the spring constant is mathematically represented  as

         k  =  \frac{4 \pi^2 * m }{ T^2}

=>      k  =  \frac{4* 3.142^2 * 1.3 }{ 1^2}

=>      k  =  51.34 \  N/ m

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Answer:

v =3.41 m/s

Explanation:

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A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi
KIM [24]

The stone reaches the wall at a height of <u>1.62 m</u>.

The stone lands at a point <u>24.5 m</u> from the point of projection.

The stone is projected horizontally with a velocity u at a height <em>h</em> from the ground. The wall is located at a distance <em>x</em> from the point of projection. The stone takes a time <em>t</em> to reach the wall and in the same time the stone falls a vertical distance <em>y</em>.

The horizontal distance <em>x</em> is traveled with a constant velocity <em>u</em>.

x=ut

Calculate the time taken <em>t</em>.

t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s

The stone's initial vertical velocity is zero. It falls through a distance <em>y</em> in the time <em>t</em> under the action of acceleration due to gravity <em>g</em>.

y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m

The height  <em>h₁ </em>of the stone above the ground when it reaches the wall  is given by,

h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m

When the stone reaches the wall, its height from the ground is <u>1.62m.</u>

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance <em>R</em> from the point of projection. If the time taken by the stone to reach the ground is <em>t₁, </em>then,

h=\frac{1}{2} gt_1^2

Calculate the time taken by the stone to reach the ground.

t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s

The horizontal distance traveled by the stone is given by,

R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

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