Red, yellow, and blue are

You attach a 3.10 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched b

dolphi86 [110]

Answer:

0.785 m/s

Explanation:

Hi!

To solve this problem we will use the equation of motion of the harmonic oscillator, i.e.

$x(t) = A cos(\omega t )+B sin(\omega t)$ - (1)

$\frac{dx}{dt}(t) = \omega (B cos(\omega t )- A sin(\omega t)$ - (1)

The problem say us that the spring is released from rest when the spring is stretched by 0.100 m, this condition is given as:

$x(0) = 0.100$

$\frac{dx}{dt}(0) = 0$

Since cos(0)=1 and sin(0) = 0:

$x(0)=A$

$\frac{dx}{dt}(0) = -B\omega$

We get

$A =0.100\\B = 0$

Now it say that after 0.4s the weigth reaches zero speed. This will happen when the sping shrinks by 0.100. This condition is written as:

$x(0.4) = - 0.100$

Since

$x(t) = 0.100 cos(\omega t)\\ -0.100=x(0.4)=0.100cos(\omega 0.4)$

This is the same as:

$-1 = cos(0.4\omega)$

We know that cosine equals to -1 when its argument is equal to:

(2n+1)π

With n an integer

The first time should happen when n=0

Therefore:

π = 0.4ω

or

ω = π/0.4 -- (2)

Now, the maximum speed will be reached when the potential energy is zero, i.e. when the sping is not stretched, that is when x = 0

With this info we will know at what time it happens:

$0 = x(t) = 0.100cos(\omega t)$

The first time that the cosine is equal to zero is when its argument is equal to π/2

i.e.

$t_{maxV}=\pi /(2\omega)$

And the velocity at that time is:

$\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\omega t_{maxV})\\\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\pi/2)\\$

But sin(π/2) = 1.

Therefore, using eq(2):

$\frac{dx}{dt}(t_{maxV} ) = 0.100*\omega = 0.100\frac{\pi}{0.400} = \pi/4$

And so:

$V_{max} = \pi / 4 =0.785$

6 0

3 years ago

Billy did an investigation to learn whether bean plants will grow if they are given salt water. Billy repeated his experiment th

Gekata [30.6K]

B should be your answer good luck!

8 0

3 years ago

Read 2 more answers

Need help ASAP

devlian [24]

Answer:

cell pohnes convert sound waves into radio waves

electromagnetic waves used in cellphone communications are called: radio waves

To send out a radio signal far and wide.. it is called: broadcast

Explanation:

5 0

3 years ago

Read 2 more answers

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

y₁ = v₀₁ t + ½ g t²

y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

t ’= (t-t₀)

y₂ = v₀₂ t ’+ ½ g t’²

y₂ = 0 + ½ g (t-t₀)²

y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

S = y₁ -y₂

S = ½ g t²– ½ g (t-t₀)²

S = ½ g [t² - (t²- 2 t to + to²)]

S = ½ g (2 t t₀ - t₀²)

S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to. The rock y has not left and the distance increases

For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂

3 0

4 years ago

Can someone help me with the exercise 3 only c,d,e please!!!

saul85 [17]

I'm not sure...
I feel ya.

4 0

3 years ago

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