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2 years ago

What's a good way to start the problem?

1 answer:

5 0

Answer: Start by writing down the problem in your own words using as much detail as possible. Be as specific as possible in your description.

Explanation:

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How does object float

IceJOKER [234]

As you may know some objects float others do not, this is because some objects contain air or oxygen. for example an orange with a peel floats because the peal contains oxygen and an orange without a peel sinks because it does not contain air at oxygen

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3 years ago

Read 2 more answers

A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 2

Tom [10]

the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:

The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

x = A cos (wt + Ф)

w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

w= $\sqrt{ \frac{200}{0.300} }$

w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

v= $\frac{dx}{dt}$

v= - A w sin (wt +Ф)

0 = -A w sin Ф

Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

4.00 = A cos 25.8t

Speed.

50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

Cos² 25.8t = $\frac{16}{A^2}$

sin² 25.8t = $\frac{3.756}{A^2 }$

1 = $\frac{1}{A^2} \ (16 + 3.756)$

A = $\sqrt{19.756}$

A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here: brainly.com/question/14311816

4 0

2 years ago

point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is

Sergio [31]

Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, $\rho_{w} = 1000\ kg//m^{3}$

Acceleration due to gravity, g = $9.8\ m/s^{2}$

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = $\frac{H + D}{2}$

h = $\frac{3 + 2.5}{2} = 2.75\ m$

Volume of water in the pool, V = $\pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}$

Mass of water, $m_{w} = \frac{\rho_{w}}{V}$

$m_{w} = \frac{1000}{311.02} = 3.215\ kg$

Work done is given by the potential energy of the water as:

$W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J$

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done, $W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J$

7 0

3 years ago

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

3 years ago

If the wavelength of a wave is quadrupied then what happens to the frequency of the wqve(Assume speed is constant)

dimulka [17.4K]

Answer:

What happens to the wavelength of a wave if you double the frequency?

If the frequency of a wave is increased, what happens to its wavelength? As the frequency increases, the wavelength decreases. 2. If the frequency is doubled, the wavelength is only half as long.

Explanation:

4 0

3 years ago