This rock appear to have formed in materials deposited in water it is most likely

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0

3 years ago

A ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec. Assuming that the air resistance can

Volgvan

Answer:

Maximum height of the ball, h(t) = 27.56 m

Explanation:

It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.

The height of the ball as a function of time t is given by :

$h(t)=h_o+v_ot-16t^2$

h₀ is initial height, h₀ = 0

So, $h(t)=42t-16t^2$ .........(1)

For maximum/minimum height, $\dfrac{dh(t)}{dt}=0$

$42-32t=0$ ...(2)

t = 1.31 s

Differentiating equation (2) wrt t

h''(t) = -32 < 0

So, at t = 1.31 seconds we will get the maximum height.

Put the value of t in equation (1)

$h(t)=42\times 1.31-16\times (1.31)^2$

h(t) = 27.56 m

Hence, this is the required solution.

7 0

4 years ago

Adrianne has just poured a cup of hot coffee. When she grabs the cup, it is very hot because the heat has transferred from the c

andrezito [222]

Answer:

A

Explanation:

because it produces heat very fast

4 0

3 years ago

Read 2 more answers

If the earth had no atmospheres at all what would the sky look like

Stells [14]

Explanation:

After the shock of seeing animals falling from the sky and silence, comes what you may have already imagined; that is, we would die from lack of oxygen - and in about 3 minutes. The worst thing is that not even patients connected to respirators in hospitals or divers equipped with cylinders could survive, since there is a need for pressure to be able to breathe.

8 0

3 years ago

In a historical movie, two knight on horseback start from

mixer [17]

Answer:

The knights collide 53.0 m from the starting point of sir George.

Explanation:

The equation for the position in a straight accelerated movement is as follows:

x = x0 + v0 t + 1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial speed

a = acceleration

t = time

The position of the two knights is the same when they collide. Since they start from rest, v0 = 0:

Sir George´s position:

xGeorge = 0 m + 0 m + 1/2 * 0.300 m/s² * t²

Considering the center of the reference system as Sir George´s initial position, the initial position of sir Alfred will be 88.0 m. The acceleration of sir Afred will be negative because he rides in opposite direction to sir George:

xAlfred = 88.0 m + 0 m - 1/2 * 0.200 m/s² * t²

When the knights collide:

xGeorge = x Alfred

1/2 * 0.300 m/s² * t² = 88.0 m - 1/2 * 0.200 m/s² * t²

0.150 m/s² * t² = 88.0 m - 0.100 m/s² * t²

0.150 m/s² * t² + 0.100 m/s² * t² = 88.0 m

0.250 m/s² * t² = 88.0 m

t² = 88.0 m / 0.250 m/s²

t = 18.8 s

At t = 18.8 s the position of sir George will be

x = 1/2 * 0.300 m/s² * (18.8 s)² = <u>53.0 m </u>

5 0

3 years ago