An ant takes 5 secs to complete one round

The magnitude of the force associated with the gravitational field is constant and has a value f. A particle is launched from po

hodyreva [135]

The particle will have an energy of 12U0. A particle travels at an initial velocity from point B to point A, gaining U0 joules of kinetic energy along the way. The constant force at this point is equal to 12F.

<h3>Does 9.8 represent gravity?</h3>

The acceleration which gravity gives to objects falling freely serves as a gauge of its strength. The gravity's acceleration at Earth's surface is approximately 9.8 meters (32 feet) per second every second.

<h3>What is a good illustration of gravity?</h3>

The following are a few instances of the power of gravity: the energy holding the gases inside the sun. the power behind a ball's descent after being thrown into the air. the force that makes an automobile coast down even when the gas is not depressed.

To know more about Gravitational visit:

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6 0

1 year ago

b) Si la distancia entre el punto A y el punto B es de 650 metros aproximadamente y la estudiante tarda 15 minutos (1200 segundo

lina2011 [118]

Answer: A
explanation:

6 0

2 years ago

In diving to a depth of 308 m, an elephant seal also moves 579 m due east of his starting point. What is the magnitude of the se

Kisachek [45]

Answer:

655.82m

Explanation:

By tracing the distances traveled from 308m down in the water and then 579m to the east, a <u>right triangle </u>can be formed and the displacement can be found by the Pythagorean theorem. This is shown in the attached image (the red line is the displacement).

One leg of the triangle will be the 308m

and the second leg of the triangle will be the 579m.

If we call the displacement $x$ , by the pythagorean theorem:

$x=\sqrt{308^2+579^2}$

$x=\sqrt{94,864+335,241}$

$x=\sqrt430,105}$

$x=655.82m$

The displacement is 655.82m

3 0

3 years ago

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, <em>i.e.</em>

$x(t) = A cos(\omega t )+B sin(\omega t)$ - (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, <em>i.e. </em>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

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

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

What is energy and types of energy<br>

pishuonlain [190]

Energy- the ability to do work/how things can change and move

Types
Potential Energy
Kinetic Energy
Nuclear Energy
Mechanical Energy
Sound Energy
Heat

8 0

2 years ago

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