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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

So

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

So

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

8 0

3 years ago

Un soldado de 1.80 m de altura ha sido herido por una bala, cuya masa es de 100 g. Deciden colocar al soldado en una centrifugad

maks197457 [2]

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3 0

3 years ago

Explain your answers to 9a and 9b in terms of Newton's laws of motion.

kogti [31]

Answer:i One way to solve the quadratic equation x2 = 9 is to subtract 9 from both sides to get one side equal to 0: x2 – 9 = 0. The expression on the left can be factored:

Explanation:

4 0

2 years ago

to start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 53.0° above the ho

kap26 [50]

So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.

The x location therefore is 290*cos(53)*35 = 6108.4m

The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.

This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m

So your (x,y) coordinates equals (6108.4, 2097.5)

5 0

3 years ago

Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo

Elena-2011 [213]

a) $3.14 \cdot 10^{-4} s$

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when $sin(...)=1$ , and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

$\lambda=2L$